JEE Complex Numbers Questions

We denote O as the origin, A by $$z_1 = \sqrt{3} + 2\sqrt{2}\,i$$, and B by $$z_2$$, where $$\sqrt{3}\,|z_2| = |z_1|$$ and $$\arg(z_2) = \arg(z_1) + \frac{\pi}{6}$$, and we aim to identify the correct statement about triangle ABO.

Since $$|z_1| = \sqrt{(\sqrt{3})^2 + (2\sqrt{2})^2} = \sqrt{3 + 8} = \sqrt{11}$$, it follows from $$\sqrt{3}\,|z_2| = \sqrt{11}$$ that $$|z_2| = \frac{\sqrt{11}}{\sqrt{3}} = \sqrt{\frac{11}{3}}\,. $$

We write $$\arg(z_1) = \tan^{-1}\!\bigl(\tfrac{2\sqrt{2}}{\sqrt{3}}\bigr) = \theta$$ (say), so that $$z_2 = |z_2|e^{i(\theta + \pi/6)} = \sqrt{\frac{11}{3}}\;e^{i(\theta + \pi/6)}\,. $$

Noting that $$OA = |z_1| = \sqrt{11}$$ and $$OB = |z_2| = \sqrt{\frac{11}{3}}\,, $$ we observe immediately that $$OA \neq OB$$, so triangle ABO is not isosceles in the sides OA and OB.

In order to compare AB and OB, we compute

$$|AB|^2 = |z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2\,|z_1|\,|z_2|\cos\Bigl(\tfrac{\pi}{6}\Bigr) = 11 + \tfrac{11}{3} - 2\sqrt{11}\,\sqrt{\tfrac{11}{3}}\;\frac{\sqrt{3}}{2} = \frac{11}{3}\,, $$

which gives $$|AB| = \sqrt{\frac{11}{3}} = |z_2| = OB\,. $$ Hence AB = OB, and triangle ABO is isosceles with AB = OB.

To determine whether it is obtuse, we apply the cosine rule at B opposite the longest side OA. Since

$$OA^2 = OB^2 + AB^2 - 2\,OB\,AB\cos B$$

we have

$$11 = \tfrac{11}{3} + \tfrac{11}{3} - 2\cdot\tfrac{11}{3}\cos B \quad\Longrightarrow\quad \cos B = -\tfrac{1}{2}\,, $$

so that $$B = 120^\circ > 90^\circ$$ and triangle ABO is obtuse.

Triangle ABO is an obtuse angled isosceles triangle, which corresponds to Option 2.

We are given $$|z| = 1$$ and $$\frac{2 + k^2z}{k + \bar{z}} = kz$$ where $$k \in \mathbb{R}$$.

Since $$|z| = 1$$, we have $$z\bar{z} = 1$$, so $$\bar{z} = \frac{1}{z}$$.

Substituting: $$\frac{2 + k^2z}{k + \frac{1}{z}} = kz$$.

Multiplying numerator and denominator of the left side by $$z$$: $$\frac{z(2 + k^2z)}{kz + 1} = kz$$.

$$\frac{2z + k^2z^2}{kz + 1} = kz$$

$$2z + k^2z^2 = kz(kz + 1) = k^2z^2 + kz$$

$$2z = kz$$

Since $$z \neq 0$$ (as $$|z| = 1$$), we get $$k = 2$$.

So the point $$k + ik^2 = 2 + 4i$$.

The circle is $$|z - (1 + 2i)| = 1$$ with center $$(1, 2)$$ and radius 1.

The distance from $$(2, 4)$$ to the center $$(1, 2)$$ is $$\sqrt{(2-1)^2 + (4-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$.

The maximum distance from the point to the circle is $$\sqrt{5} + 1$$ (distance to center plus radius).

Hence, the correct answer is Option A.

Let $$z = x + iy$$, where $$x, y \in \mathbb{R}$$.

Define $$w = \dfrac{z-1}{2z+i}\,.$$
Its complex conjugate is $$\overline{w}= \dfrac{\bar z-1}{2\bar z-i}\,.$

The given condition is
$$$$\text{Re}$$\!$$\left$$(w$$\right$$)+$$\text{Re}$$\!$$\left$$($$\overline{w}$$$$\right$$)=2.$$

For any complex number $$w$$, $$$$\text{Re}$$(w)=$$\text{Re}$$($$\overline{w}$$)$$. Therefore
$$2\,$$\text{Re}$$(w)=2 \;\Longrightarrow\; $$\text{Re}$$(w)=1.$$

Hence we need the locus of $$z$$ satisfying
$$$$\text{Re}$$\!$$\left$$(\dfrac{z-1}{2z+i}$$\right$$)=1.$$

Write numerator and denominator in terms of $$x, y$$:
$$z-1=(x-1)+iy,$$
$$2z+i=2x+i(2y+1).$$

Multiply by the conjugate of the denominator to extract the real part:
$$\dfrac{z-1}{2z+i}= \dfrac{(x-1)+iy}{2x+i(2y+1)}\; $$\cdot$$\dfrac{2x-i(2y+1)}{2x-i(2y+1)}.$$

Denominator magnitude squared:
$$D=(2x)^2+(2y+1)^2=4x^2+(2y+1)^2.$$

Numerator product:
$$\;(x-1)+iy$$$$\cdot$$$$2x-i(2y+1)$$ gives

Real part $$R=2x(x-1)+y(2y+1)=2x^2-2x+2y^2+y.$$

Thus
$$$$\text{Re}$$\!$$\left$$(\dfrac{z-1}{2z+i}$$\right$$)=\dfrac{R}{D}=1 \;\Longrightarrow\; R=D.$$

Equate and rearrange:
$$2x^2-2x+2y^2+y =4x^2+4y^2+4y+1,$$
$$0=2x^2+2y^2+2x+3y+1.$$

Divide by 2 for simplicity:
$$x^2+y^2+x+\dfrac{3}{2}y+\dfrac{1}{2}=0.$$

Complete the squares:

$$x^2+x=$$\left$$(x+\dfrac{1}{2}$$\right$$)^2-\dfrac{1}{4},$$
$$y^2+\dfrac{3}{2}y=$$\left$$(y+\dfrac{3}{4}$$\right$$)^2-$$\left$$(\dfrac{3}{4}$$\right$$)^2.$$

Substitute:

$$$$\left$$(x+\dfrac{1}{2}$$\right$$)^2-\dfrac{1}{4} +$$\left$$(y+\dfrac{3}{4}$$\right$$)^2-\dfrac{9}{16} +\dfrac{1}{2}=0.$$

Combine constants:
$$-\dfrac{1}{4}-\dfrac{9}{16}+\dfrac{1}{2} =-\dfrac{5}{16}.$$

Move to the right side:

$$$$\left$$(x+\dfrac{1}{2}$$\right$$)^2+$$\left$$(y+\dfrac{3}{4}$$\right$$)^2 =\dfrac{5}{16}.$$

Hence the locus is a circle with
center $$\bigl(a,b\bigr)=$$\left$$(-\dfrac{1}{2},\,-\dfrac{3}{4}$$\right$$),$$
radius $$r=$$\sqrt{\dfrac{5}{16}$$}=\dfrac{$$\sqrt{5}$$}{4},$$
so $$r^{2}=\dfrac{5}{16}.$$

Now evaluate $$\dfrac{15ab}{r^{2}}.$$
$$ab=$$\left$$(-\dfrac{1}{2}$$\right$$)\!$$\left$$(-\dfrac{3}{4}$$\right$$)=\dfrac{3}{8},$$
$$\dfrac{15ab}{r^{2}}=\dfrac{15$$\cdot$$$$\frac{3}{8}$$}{$$\frac{5}{16}$$} =\dfrac{45}{8}$$\cdot$$\dfrac{16}{5}=18.$$

The required value is $$18$$, which corresponds to Option C.

We need to find the minimum value of $$|z_1 - z_2|$$ given $$|z_1 - 8 - 2i| \leq 1$$ and $$|z_2 - 2 + 6i| \leq 2$$. Here, $$z_1$$ lies in or on the circle centered at $$(8,2)$$ with radius 1, and $$z_2$$ lies in or on the circle centered at $$(2,-6)$$ with radius 2. The distance between these centers is $$d = \sqrt{(8-2)^2 + (2-(-6))^2} = \sqrt{36 + 64} = \sqrt{100} = 10.$$ Therefore, the minimum possible distance between any point in the first circle and any point in the second circle is $$d - 1 - 2 = 10 - 1 - 2 = 7.$$

The correct answer is Option 4: 7.

The non-real cube roots of unity are $$\omega$$ and $$\omega^{2}$$, where $$\omega^{3}=1$$ and $$1+\omega+\omega^{2}=0$$.
Because $$0 \lt \arg(\omega) \lt \pi$$, we take $$\omega = e^{\,i\frac{2\pi}{3}} = -\frac12 + i\frac{\sqrt3}{2}$$, so $$\arg(\omega)=\frac{2\pi}{3}$$.

Define $$r=-\omega$$. Then
$$r = -\omega = e^{\,i\pi}\,e^{\,i\frac{2\pi}{3}} = e^{\,i\frac{5\pi}{3}} = e^{-\,i\frac{\pi}{3}}$$
and hence $$\arg(r)=-\frac{\pi}{3}$$.

Since $$r^{3}=(-\omega)^{3}=-(\omega^{3})=-1$$, we get $$r^{6}=1$$. Thus the powers of $$r$$ repeat every six terms.

The required sum is $$S=\sum_{n=1}^{2025} r^{\,n}.$$ Because one full cycle of six consecutive terms sums to zero (geometric-series property for a 6th root of unity different from 1), we split 2025 terms into cycles:

$$2025 = 6 \times 337 + 3.$$ The first $$6\times337=2022$$ terms contribute zero, so $$S = r^{2023}+r^{2024}+r^{2025}.$$

Now reduce exponents modulo 6: $$2023 \equiv 1,\; 2024 \equiv 2,\; 2025 \equiv 3 \pmod{6}.$$ Hence $$S = r^{1}+r^{2}+r^{3}.$$

Compute each term:
$$r^{1} = -\omega,$$ $$r^{2} = (-\omega)^{2} = \omega^{2},$$ $$r^{3} = -1.$$ Therefore $$S = -\omega + \omega^{2} - 1.$$

Using $$1+\omega+\omega^{2}=0 \; \Rightarrow \; \omega^{2}= -1-\omega,$$ $$S = -\omega + (-1-\omega) -1 = -2 - 2\omega = -2(1+\omega).$$

But $$1+\omega = -\omega^{2},$$ so $$S = -2(-\omega^{2}) = 2\,\omega^{2}.$$

Multiplying by the positive real number 2 does not change the argument, hence $$\alpha = \arg(S) = \arg(\omega^{2}).$$ For our chosen branch, $$\omega^{2}=e^{\,i\frac{4\pi}{3}}$$ has principal argument $$\frac{4\pi}{3}-2\pi = -\frac{2\pi}{3}.$$ Thus $$\alpha = -\frac{2\pi}{3}.$$

Finally, $$\frac{3\alpha}{\pi} = \frac{3\left(-\tfrac{2\pi}{3}\right)}{\pi} = -2.$$

Answer: -2

Write $$\cos\theta = c$$ and $$\sin\theta = s$$ for brevity.

The given condition is
$$1 + 10\,\text{Re}\!\left(\dfrac{2c + i\,s}{\,c - 3i\,s}\right)=0$$
which can be rearranged as
$$\text{Re}\!\left(\dfrac{2c + i\,s}{\,c - 3i\,s}\right)= -\dfrac1{10}\,\,\,\,\,\,-(1)$$

To extract the real part, multiply numerator and denominator by the conjugate of the denominator:

$$\dfrac{2c + i\,s}{\,c - 3i\,s} =\dfrac{(2c + i\,s)(c + 3i\,s)}{(c - 3i\,s)(c + 3i\,s)} =\dfrac{(2c + i\,s)(c + 3i\,s)}{c^2 + 9s^2}$$

Expand the numerator:
$$\begin{aligned} (2c + i\,s)(c + 3i\,s) &= 2c\cdot c + 2c\cdot 3i\,s + i\,s\cdot c + i\,s\cdot 3i\,s\\ &= 2c^2 + 6i\,cs + i\,cs + 3i^2s^2\\ &= 2c^2 + 7i\,cs - 3s^2 \end{aligned}$$

Thus
$$\dfrac{2c + i\,s}{\,c - 3i\,s} =\dfrac{\,2c^2 - 3s^2}{c^2 + 9s^2} + i\,\dfrac{7cs}{c^2 + 9s^2}$$

Hence the real part is
$$\text{Re} = \dfrac{\,2c^2 - 3s^2}{c^2 + 9s^2}$$

Insert this into $$(1)$$:
$$\dfrac{\,2c^2 - 3s^2}{c^2 + 9s^2} = -\dfrac1{10}$$

Cross-multiply:
$$10(2c^2 - 3s^2) = -(c^2 + 9s^2)$$
$$20c^2 - 30s^2 + c^2 + 9s^2 = 0$$
$$21c^2 - 21s^2 = 0$$
$$c^2 = s^2$$

Therefore $$|\,\cos\theta| = |\,\sin\theta|$$, which is equivalent to
$$\tan^2\theta = 1 \; \Longrightarrow \; \theta = \dfrac{\pi}{4} + k\dfrac{\pi}{2}, \; k \in \mathbb{Z}$$

Within the interval $$[0, 2\pi]$$ the admissible angles are
$$\theta_1 = \dfrac{\pi}{4},\; \theta_2 = \dfrac{3\pi}{4},\; \theta_3 = \dfrac{5\pi}{4},\; \theta_4 = \dfrac{7\pi}{4}$$

Compute the required sum:
$$\sum_{\theta\in A} \theta^2 = \left(\dfrac{\pi}{4}\right)^2 + \left(\dfrac{3\pi}{4}\right)^2 + \left(\dfrac{5\pi}{4}\right)^2 + \left(\dfrac{7\pi}{4}\right)^2$$
$$= \dfrac{\pi^2}{16} + \dfrac{9\pi^2}{16} + \dfrac{25\pi^2}{16} + \dfrac{49\pi^2}{16} = \dfrac{84\pi^2}{16} = \dfrac{21}{4}\pi^2$$

Hence $$\displaystyle \sum_{\theta \in A} \theta^2 = \dfrac{21}{4}\pi^2$$, which matches Option A.

The given condition is
$$\frac{z^{2}+3i}{z-2+i}=2+3i$$.

First cross-multiply:
$$z^{2}+3i=(2+3i)\,(z-2+i).$$

Expand the right-hand side.
We need $$(2+3i)(z-2+i)=(2+3i)z+(2+3i)(-2+i).$$

Compute the constant product:
$$(2+3i)(-2+i)=-4-6i+2i+3i^{2}=-4-4i-3=-7-4i.$$

Hence
$$z^{2}+3i=(2+3i)z-7-4i.$$

Bring all terms to the left:
$$z^{2}-(2+3i)z+(3i+7+4i)=0.$$

Simplify the constant term $$3i+4i=7i$$, so the quadratic in $$z$$ is
$$z^{2}-(2+3i)z+(7+7i)=0.$$

For a quadratic $$z^{2}+Bz+C=0$$ with roots $$z_{1},z_{2}$$:
• Sum of roots $$S=z_{1}+z_{2}=-B.$$
• Product of roots $$P=z_{1}z_{2}=C.$$

Here $$B=-(2+3i)$$ and $$C=7+7i$$, hence
$$S=2+3i,\qquad P=7+7i.$$

We want the sum of all possible values of $$z^{2}$$, namely $$z_{1}^{2}+z_{2}^{2}$$.
Use the identity
$$z_{1}^{2}+z_{2}^{2}=(z_{1}+z_{2})^{2}-2z_{1}z_{2}=S^{2}-2P.$$

Compute $$S^{2}$$:
$$(2+3i)^{2}=4+12i+9i^{2}=4+12i-9=-5+12i.$$

Compute $$2P$$:
$$2P=2(7+7i)=14+14i.$$

Therefore
$$z_{1}^{2}+z_{2}^{2}=(-5+12i)-(14+14i)=-19-2i.$$

Thus the sum of all possible values of $$z^{2}$$ is $$-19-2i$$, which matches Option B.

We need to find the number of complex numbers z satisfying $$|z| = 1$$ and $$\left|\frac{z}{\bar{z}} + \frac{\bar{z}}{z}\right| = 1$$.

Let $$z = e^{i\theta}$$

Since $$|z| = 1$$, we have $$\bar{z} = e^{-i\theta}$$.

$$\frac{z}{\bar{z}} = e^{2i\theta}$$ and $$\frac{\bar{z}}{z} = e^{-2i\theta}$$

$$\frac{z}{\bar{z}} + \frac{\bar{z}}{z} = e^{2i\theta} + e^{-2i\theta} = 2\cos 2\theta$$

Apply the condition

$$|2\cos 2\theta| = 1$$

$$\cos 2\theta = \pm \frac{1}{2}$$

Solve for $$\theta$$

Case 1: $$\cos 2\theta = \frac{1}{2}$$

$$2\theta = \pm \frac{\pi}{3} + 2n\pi$$

$$\theta = \pm \frac{\pi}{6} + n\pi$$

In $$[0, 2\pi)$$: $$\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$$ — 4 values

Case 2: $$\cos 2\theta = -\frac{1}{2}$$

$$2\theta = \pm \frac{2\pi}{3} + 2n\pi$$

$$\theta = \pm \frac{\pi}{3} + n\pi$$

In $$[0, 2\pi)$$: $$\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$$ — 4 values

Total count

Total = 4 + 4 = 8 complex numbers.

The correct answer is Option 2: 8.

Given that $$z_1, z_2, z_3$$ are vertices of an equilateral triangle with centroid $$z_0$$.

The centroid is $$z_0 = \dfrac{z_1 + z_2 + z_3}{3}$$.

For an equilateral triangle with centroid at $$z_0$$, the vertices can be written as $$z_k = z_0 + r \cdot e^{i(\theta + 2\pi(k-1)/3)}$$ for $$k = 1, 2, 3$$, where $$r$$ is the circumradius.

So $$z_k - z_0 = r \cdot e^{i(\theta + 2\pi(k-1)/3)}$$.

Therefore: $$\displaystyle\sum_{k=1}^{3} (z_k - z_0)^2 = r^2 \displaystyle\sum_{k=1}^{3} e^{2i(\theta + 2\pi(k-1)/3)}$$

$$= r^2 e^{2i\theta} \displaystyle\sum_{k=1}^{3} e^{4\pi i(k-1)/3}$$

$$= r^2 e^{2i\theta} \left(1 + e^{4\pi i/3} + e^{8\pi i/3}\right)$$

Now, $$e^{4\pi i/3}$$ and $$e^{8\pi i/3} = e^{2\pi i/3}$$ (since $$8\pi/3 - 2\pi = 2\pi/3$$) are the cube roots of unity (other than 1 shifted). Specifically, $$1 + e^{2\pi i/3} + e^{4\pi i/3} = 0$$ (sum of cube roots of unity).

Therefore, $$\displaystyle\sum_{k=1}^{3} (z_k - z_0)^2 = r^2 e^{2i\theta} \cdot 0 = 0$$.

Hence, the correct answer is Option A.

Given $$\left|\frac{\bar{z}-i}{2\bar{z}+i}\right| = \frac{1}{3}$$, find $$\alpha^2$$ where the area of the triangle with vertices $$(0,0)$$, center $$C$$, and $$(\alpha, 0)$$ is 11.

Let $$z = x + iy$$, so $$\bar{z} = x - iy$$.

$$\left|\frac{(x-iy) - i}{2(x-iy) + i}\right| = \frac{1}{3}$$

$$\left|\frac{x - i(y+1)}{2x - i(2y-1)}\right| = \frac{1}{3}$$

Squaring and using $$|w_1/w_2|^2 = |w_1|^2/|w_2|^2$$:

$$\frac{x^2 + (y+1)^2}{4x^2 + (2y-1)^2} = \frac{1}{9}$$

$$9[x^2 + (y+1)^2] = 4x^2 + (2y-1)^2$$

$$9x^2 + 9y^2 + 18y + 9 = 4x^2 + 4y^2 - 4y + 1$$

$$5x^2 + 5y^2 + 22y + 8 = 0$$

$$x^2 + y^2 + \frac{22}{5}y + \frac{8}{5} = 0$$

Completing the square: $$x^2 + \left(y + \frac{11}{5}\right)^2 = \frac{121}{25} - \frac{8}{5} = \frac{121 - 40}{25} = \frac{81}{25}$$

Center $$C = \left(0, -\frac{11}{5}\right)$$, radius $$= \frac{9}{5}$$.

Vertices: $$(0,0)$$, $$\left(0, -\frac{11}{5}\right)$$, $$(\alpha, 0)$$.

Area = $$\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

$$= \frac{1}{2}\left|0 + 0 + \alpha\left(0 + \frac{11}{5}\right)\right| = \frac{11|\alpha|}{10}$$

Setting this equal to 11: $$\frac{11|\alpha|}{10} = 11 \implies |\alpha| = 10$$

$$\alpha^2 = 100$$

The correct answer is Option 2: 100.

We need to find $$\alpha\gamma + \beta\delta$$ where $$\alpha + i\beta$$ and $$\gamma + i\delta$$ are roots of $$x^2 - (3-2i)x - (2i-2) = 0$$.

We start by using Vieta’s formulas: the sum of roots $$(\alpha + i\beta) + (\gamma + i\delta) = 3 - 2i$$ and the product of roots $$(\alpha + i\beta)(\gamma + i\delta) = -(2i - 2) = 2 - 2i$$.

Next, expanding the product gives $$(\alpha + i\beta)(\gamma + i\delta) = \alpha\gamma - \beta\delta + i(\alpha\delta + \beta\gamma) = 2 - 2i$$ which leads to the real and imaginary parts $$\alpha\gamma - \beta\delta = 2$$ and $$\alpha\delta + \beta\gamma = -2$$.

In order to determine $$\alpha\gamma + \beta\delta$$, we apply the quadratic formula: $$x = \frac{(3-2i) \pm \sqrt{(3-2i)^2 + 4(2i-2)}}{2}$$.

We note $$(3-2i)^2 = 9 - 12i + 4i^2 = 5 - 12i$$ and thus the discriminant is $$\Delta = 5 - 12i + 8i - 8 = -3 - 4i$$.

Substituting $$\sqrt{-3-4i} = a + bi$$ yields the system $$a^2 - b^2 = -3$$ and $$2ab = -4$$. From $$2ab = -4$$ we have $$b = -2/a$$, and substituting into $$a^2 - b^2 = -3$$ gives $$a^4 + 3a^2 - 4 = 0$$.

Factoring as $$(a^2 + 4)(a^2 - 1) = 0$$ implies $$a = \pm 1$$. If $$a = 1$$ then $$b = -2$$, so $$\sqrt{-3-4i} = 1 - 2i$$ (up to sign).

Therefore, the roots are $$x = \frac{(3-2i) \pm (1-2i)}{2}$$, giving $$x_1 = \frac{4-4i}{2} = 2 - 2i$$ and $$x_2 = \frac{2}{2} = 1$$.

Hence one root is $$\alpha + i\beta = 2 - 2i$$ and the other is $$\gamma + i\delta = 1 + 0i$$ (or vice versa).

Substituting into $$\alpha\gamma + \beta\delta$$ yields $$2 \cdot 1 + (-2) \cdot 0 = 2$$.

Therefore, the correct answer is Option D) 2.

The given curve is $$z(1+i) + \overline{z}(1-i) = 4$$, where $$z \in \mathbb{C}$$. Let $$z = x + iy$$, so $$\overline{z} = x - iy$$. Substituting into the equation:

$$ (x + iy)(1 + i) + (x - iy)(1 - i) = 4 $$

Expanding each term:

$$ (x + iy)(1 + i) = x \cdot 1 + x \cdot i + iy \cdot 1 + iy \cdot i = x + ix + iy + i^2 y = x + ix + iy - y = (x - y) + i(x + y) $$

$$ (x - iy)(1 - i) = x \cdot 1 + x \cdot (-i) + (-iy) \cdot 1 + (-iy) \cdot (-i) = x - ix - iy + i^2 y = x - ix - iy - y = (x - y) - i(x + y) $$

Adding the terms:

$$ [(x - y) + i(x + y)] + [(x - y) - i(x + y)] = (x - y + x - y) + i(x + y - x - y) = 2x - 2y $$

Setting equal to 4:

$$ 2x - 2y = 4 \implies x - y = 2 $$

Thus, the curve is the straight line $$x - y = 2$$.

The region $$|z - 3| \leq 1$$ is a disk centered at $$(3, 0)$$ with radius 1, since $$|(x - 3) + iy| \leq 1$$ gives $$(x - 3)^2 + y^2 \leq 1$$.

To find how the line $$x - y = 2$$ divides this disk, first find the points of intersection by substituting $$y = x - 2$$ into the circle equation:

$$ (x - 3)^2 + (x - 2)^2 = 1 $$

Expanding:

$$ (x^2 - 6x + 9) + (x^2 - 4x + 4) = 1 \implies 2x^2 - 10x + 13 = 1 \implies 2x^2 - 10x + 12 = 0 $$

Dividing by 2:

$$ x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0 $$

So $$x = 2$$ or $$x = 3$$. When $$x = 2$$, $$y = 2 - 2 = 0$$; when $$x = 3$$, $$y = 3 - 2 = 1$$. The intersection points are $$(2, 0)$$ and $$(3, 1)$$.

The chord joining $$(2, 0)$$ and $$(3, 1)$$ subtends an angle at the center $$(3, 0)$$. The vectors from the center to these points are $$\overrightarrow{CA} = (-1, 0)$$ and $$\overrightarrow{CB} = (0, 1)$$. The dot product is $$(-1)(0) + (0)(1) = 0$$, and the magnitudes are both 1, so $$\cos \theta = 0$$, giving $$\theta = \pi/2$$ radians.

The disk has area $$\pi r^2 = \pi \cdot 1^2 = \pi$$. The chord divides the disk into two segments: a minor segment and a major segment. The area of a segment is given by $$\frac{r^2}{2} (\theta - \sin \theta)$$, where $$\theta$$ is the central angle in radians.

For the minor segment, $$\theta = \pi/2$$:

$$ \text{Area} = \frac{1}{2} \left( \frac{\pi}{2} - \sin \frac{\pi}{2} \right) = \frac{1}{2} \left( \frac{\pi}{2} - 1 \right) = \frac{\pi}{4} - \frac{1}{2} $$

For the major segment, $$\theta = 2\pi - \pi/2 = 3\pi/2$$:

$$ \text{Area} = \frac{1}{2} \left( \frac{3\pi}{2} - \sin \frac{3\pi}{2} \right) = \frac{1}{2} \left( \frac{3\pi}{2} - (-1) \right) = \frac{1}{2} \left( \frac{3\pi}{2} + 1 \right) = \frac{3\pi}{4} + \frac{1}{2} $$

Let $$\alpha$$ be the minor segment area and $$\beta$$ the major segment area:

$$ \alpha = \frac{\pi}{4} - \frac{1}{2}, \quad \beta = \frac{3\pi}{4} + \frac{1}{2} $$

The absolute difference is:

$$ |\alpha - \beta| = \left| \left( \frac{\pi}{4} - \frac{1}{2} \right) - \left( \frac{3\pi}{4} + \frac{1}{2} \right) \right| = \left| \frac{\pi}{4} - \frac{1}{2} - \frac{3\pi}{4} - \frac{1}{2} \right| = \left| -\frac{2\pi}{4} - 1 \right| = \left| -\frac{\pi}{2} - 1 \right| = \frac{\pi}{2} + 1 $$

Since $$\beta > \alpha$$, $$\beta - \alpha = \left( \frac{3\pi}{4} + \frac{1}{2} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{2\pi}{4} + 1 = \frac{\pi}{2} + 1$$, confirming the result.

The value $$\frac{\pi}{2} + 1$$ matches option A: $$1 + \frac{\pi}{2}$$.

Thus, $$|\alpha - \beta| = 1 + \frac{\pi}{2}$$.

Given three complex numbers on the unit circle $$|z| = 1$$ with arguments $$\arg(z_1) = -\frac{\pi}{4}$$, $$\arg(z_2) = 0$$, and $$\arg(z_3) = \frac{\pi}{4}$$.

Since they lie on the unit circle, they can be expressed in exponential form as:

$$z_1 = e^{-i\pi/4}, \quad z_2 = e^{i \cdot 0} = 1, \quad z_3 = e^{i\pi/4}$$

The conjugates are:

$$\overline{z_1} = e^{i\pi/4}, \quad \overline{z_2} = e^{-i \cdot 0} = 1, \quad \overline{z_3} = e^{-i\pi/4}$$

We need to compute the expression:

$$S = z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}$$

Substituting the values:

$$z_1 \overline{z_2} = e^{-i\pi/4} \cdot 1 = e^{-i\pi/4}$$

$$z_2 \overline{z_3} = 1 \cdot e^{-i\pi/4} = e^{-i\pi/4}$$

$$z_3 \overline{z_1} = e^{i\pi/4} \cdot e^{i\pi/4} = e^{i(\pi/4 + \pi/4)} = e^{i\pi/2} = i$$

Adding these together:

$$S = e^{-i\pi/4} + e^{-i\pi/4} + i = 2e^{-i\pi/4} + i$$

Now, $$e^{-i\pi/4} = \cos(-\pi/4) + i \sin(-\pi/4) = \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} (1 - i)$$. Therefore:

$$2e^{-i\pi/4} = 2 \cdot \frac{1}{\sqrt{2}} (1 - i) = \sqrt{2} (1 - i) = \sqrt{2} - i\sqrt{2}$$

So,

$$S = (\sqrt{2} - i\sqrt{2}) + i = \sqrt{2} + i(1 - \sqrt{2})$$

To find $$|S|^2$$, recall that for a complex number $$z = x + iy$$, $$|z|^2 = x^2 + y^2$$. Here, $$x = \sqrt{2}$$ and $$y = 1 - \sqrt{2}$$:

$$|S|^2 = (\sqrt{2})^2 + (1 - \sqrt{2})^2 = 2 + \left(1 - 2\sqrt{2} + (\sqrt{2})^2\right) = 2 + \left(1 - 2\sqrt{2} + 2\right) = 2 + (3 - 2\sqrt{2}) = 5 - 2\sqrt{2}$$

The expression $$|S|^2 = \alpha + \beta \sqrt{2}$$ is given, so comparing:

$$5 - 2\sqrt{2} = \alpha + \beta \sqrt{2} \implies \alpha = 5, \quad \beta = -2$$

Now, compute $$\alpha^2 + \beta^2$$:

$$\alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29$$

Thus, the value is 29.

The complex numbers in set $$A$$ satisfy the circle equation
$$|z-(2+i)|=3 \; \Longrightarrow \; (x-2)^2+(y-1)^2=9$$ where $$z=x+iy$$.

The numbers in set $$B$$ satisfy
$$\text{Re}(z-iz)=2.$$ First write $$z-iz=z(1-i)=(x+iy)(1-i).$$

Expanding,
$$(x+iy)(1-i)=x+iy-ix-i^2y=(x+y)+i(y-x).$$
Hence $$\text{Re}(z-iz)=x+y,$$ so $$B$$ is the straight line
$$x+y=2.$$

To find the intersection $$S=A\cap B,$$ solve the system
$$\begin{cases}(x-2)^2+(y-1)^2=9 \\[4pt] x+y=2\end{cases}$$

From the line, $$y=2-x.$$ Substitute into the circle:

$$(x-2)^2+\bigl((2-x)-1\bigr)^2=9\\ (x-2)^2+(1-x)^2=9.$$

Expand and collect terms:
$$(x^2-4x+4)+(x^2-2x+1)=9\\ 2x^2-6x+5=9\\ 2x^2-6x-4=0\\ x^2-3x-2=0.$$

Solve the quadratic:
$$x=\frac{3\pm\sqrt{9+8}}{2}=\frac{3\pm\sqrt{17}}{2}.$$

Using $$y=2-x$$, we get the two intersection points

$$z_1=\frac{3+\sqrt{17}}{2}+i\,\frac{1-\sqrt{17}}{2},\qquad z_2=\frac{3-\sqrt{17}}{2}+i\,\frac{1+\sqrt{17}}{2}.$$

Now compute $$|z|^2=x^2+y^2$$ for each point.

For $$z_1$$: $$x_1=\frac{3+\sqrt{17}}{2},\;y_1=\frac{1-\sqrt{17}}{2}$$ $$|z_1|^2=\left(\frac{3+\sqrt{17}}{2}\right)^2+\left(\frac{1-\sqrt{17}}{2}\right)^2 =\frac{(3+\sqrt{17})^2+(1-\sqrt{17})^2}{4} =\frac{26+6\sqrt{17}+18-2\sqrt{17}}{4} =11+\sqrt{17}.$$

For $$z_2$$: $$x_2=\frac{3-\sqrt{17}}{2},\;y_2=\frac{1+\sqrt{17}}{2}$$ $$|z_2|^2=\left(\frac{3-\sqrt{17}}{2}\right)^2+\left(\frac{1+\sqrt{17}}{2}\right)^2 =11-\sqrt{17}.$$

Finally,
$$\sum_{z\in S}|z|^2=(11+\sqrt{17})+(11-\sqrt{17})=22.$$

Therefore the required sum is $$22$$.

We have $$\alpha, \beta$$ as roots of $$x^2 - ax - b = 0$$. Given $$P_n = \alpha^n - \beta^n$$, with $$P_3 = -5\sqrt{7}i$$, $$P_4 = -3\sqrt{7}i$$, $$P_5 = 11\sqrt{7}i$$, $$P_6 = 45\sqrt{7}i$$.

Since $$\alpha, \beta$$ satisfy $$x^2 = ax + b$$, the sequence $$P_n$$ satisfies the recurrence relation

$$P_n = aP_{n-1} + bP_{n-2}$$

Substituting $$n=5$$ gives

$$P_5 = aP_4 + bP_3$$

Therefore,

$$11\sqrt{7}i = a(-3\sqrt{7}i) + b(-5\sqrt{7}i)$$

$$11 = -3a - 5b$$ ... (i)

For $$n=6$$,

$$P_6 = aP_5 + bP_4$$

Thus,

$$45\sqrt{7}i = a(11\sqrt{7}i) + b(-3\sqrt{7}i)$$

$$45 = 11a - 3b$$ ... (ii)

From (i): $$3a + 5b = -11$$ ... (i')

From (ii): $$11a - 3b = 45$$ ... (ii')

Multiply (i') by 3: $$9a + 15b = -33$$

Multiply (ii') by 5: $$55a - 15b = 225$$

Adding: $$64a = 192 \Rightarrow a = 3$$

From (i'): $$9 + 5b = -11 \Rightarrow b = -4$$

We know: $$\alpha + \beta = a = 3$$ and $$\alpha\beta = -b = 4$$

$$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 9 - 8 = 1$$

$$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = 1 - 2(16) = 1 - 32 = -31$$

$$|\alpha^4 + \beta^4| = |-31| = 31$$

The answer is 31.

The integers $$a,b$$ are chosen from $$[-3,3]$$ with the restriction $$a+b \neq 0$$.
For such a pair we have to find a complex number $$z$$ that satisfies both

$$\left|\frac{z-a}{z+b}\right|=1 \qquad -(1)$$
and

$$\begin{vmatrix} z+1 & \omega & \omega^{2}\\ \omega & z+\omega^{2} & 1\\ \omega^{2} & 1 & z+\omega \end{vmatrix}=1, \qquad -(2)$$
where $$\omega,\omega^{2}$$ are the non-real cube roots of unity, i.e. $$\omega^{2}+\omega+1=0$$ and $$\omega^{3}=1$$.

Step 1: Evaluate the determinant in (2).

Let $$D(z)$$ denote the determinant. Expanding along the first row,

$$D(z)=(z+1)\!\begin{vmatrix}z+\omega^{2}&1\\1&z+\omega\end{vmatrix}-\omega\!\begin{vmatrix}\omega&1\\\omega^{2}&z+\omega\end{vmatrix}+\omega^{2}\!\begin{vmatrix}\omega&z+\omega^{2}\\\omega^{2}&1\end{vmatrix}.$$

Compute the three $$2\times 2$$ minors:

$$\begin{aligned} A&=(z+\omega^{2})(z+\omega)-1\\ &=z^{2}+z(\omega+\omega^{2})+\omega\omega^{2}-1\\ &=z^{2}-z+1-1 \quad(\text{because } \omega+\omega^{2}=-1,\;\omega\omega^{2}=1)\\ &=z^{2}-z,\\[6pt] B&=\omega(z+\omega)-\omega^{2}=\omega z+\omega^{2}-\omega^{2}=\omega z,\\[6pt] C&=\omega-\omega^{2}(z+\omega^{2})=\omega-\omega^{2}z-\omega^{4} =\omega-\omega^{2}z-\omega=\!-\,\omega^{2}z. \end{aligned}$$

Hence

$$\begin{aligned} D(z)&=(z+1)(z^{2}-z)-\omega(\omega z)+\omega^{2}(-\omega^{2}z)\\ &=z^{3}-z-\omega^{2}z-\omega z\\ &=z^{3}-z(1+\omega+\omega^{2})\\ &=z^{3}\quad\bigl(1+\omega+\omega^{2}=0\bigr). \end{aligned}$$

Equation (2) therefore gives

$$z^{3}=1 \quad\Longrightarrow\quad z\in\{1,\;\omega,\;\omega^{2}\}.$$(3)

Step 2: Interpret the modulus condition (1).

Because $$a,b$$ are real, points $$a$$ and $$-b$$ lie on the real axis. Writing $$z=x+iy,$$

$$\left|\frac{z-a}{z+b}\right|=1 \;\Longrightarrow\;|z-a|=|z+b|,$$

which is the set of points equidistant from $$a$$ and $$-b$$. The locus is the perpendicular bisector of the segment joining $$a$$ and $$-b$$, i.e. the vertical line

$$x=\frac{a-b}{2}. \qquad -(4)$$

Step 3: Make the locus pass through at least one root of (3).

The three cube roots of unity and their real parts are

$$1 \;(x=1),\qquad \omega=-\tfrac12+\tfrac{\sqrt3}2 i \;(x=-\tfrac12),\qquad \omega^{2}=-\tfrac12-\tfrac{\sqrt3}2 i \;(x=-\tfrac12).$$

Thus a root from (3) lies on the line (4) iff

$$\frac{a-b}{2}=1 \quad\text{or}\quad \frac{a-b}{2}=-\frac12.$$

Hence we need

$$a-b=2 \qquad\text{or}\qquad a-b=-1.\quad -(5)$$

Step 4: List all integer pairs in $$[-3,3]$$ satisfying (5) and $$a+b\neq0$$.

Case 1: $$a-b=2.$$ Put $$b=a-2$$ and keep $$a,b\in[-3,3].$$

Possible $$a$$ values: $$-1,0,1,2,3.$$ Corresponding pairs:

$$(a,b)=(-1,-3),(0,-2),(1,-1),(2,0),(3,1).$$

Reject the pair with $$a+b=0$$, i.e. $$(1,-1).$$ Valid pairs: 4.

Case 2: $$a-b=-1.$$ Put $$b=a+1$$ and keep $$a,b\in[-3,3].$$

Possible $$a$$ values: $$-3,-2,-1,0,1,2.$$ Pairs:

$$(a,b)=(-3,-2),(-2,-1),(-1,0),(0,1),(1,2),(2,3).$$

All have $$a+b\neq0$$, so all 6 are valid.

Step 5: Total number of ordered pairs.

Total valid pairs $$=4+6=10.$$

Answer : 10

Write the two complex numbers as
$$\omega_1 =(8+i)\sin\theta +(7+4i)\cos\theta ,\qquad \omega_2 =(1+8i)\sin\theta +(4+7i)\cos\theta .$$

Their product is
$$\omega_1\omega_2 =(8+i)\!(1+8i)\sin^2\theta +\big[(8+i)(4+7i)+(7+4i)(1+8i)\big]\sin\theta\cos\theta +(7+4i)(4+7i)\cos^2\theta .$$

Compute the four required products one by one.

1. $$AC=(8+i)(1+8i)=65i.$$

2. $$AD=(8+i)(4+7i)=25+60i.$$

3. $$BC=(7+4i)(1+8i)=-25+60i.$$

4. $$BD=(7+4i)(4+7i)=65i.$$

Hence
$$AD+BC=(25+60i)+(-25+60i)=120i.$$

Substitute back:

$$\omega_1\omega_2 =65i\sin^2\theta+120i\sin\theta\cos\theta+65i\cos^2\theta.$$

Factor $$65i$$ in the first and last terms and use $$\sin^2\theta+\cos^2\theta=1$$:

$$\omega_1\omega_2 =65i+120i\sin\theta\cos\theta.$$

Convert the mixed term with $$\sin2\theta=2\sin\theta\cos\theta$$:

$$\omega_1\omega_2 =65i+60i\sin2\theta.$$

Therefore the product is purely imaginary. Write
$$\omega_1\omega_2=\alpha+i\beta,$$ so that $$\alpha=0,\quad \beta=65+60\sin2\theta.$$

The required sum is $$\alpha+\beta=\beta=65+60\sin2\theta.$$

Because $$-1\le\sin2\theta\le1,$$ the maximum and minimum values are

$$p=65+60(1)=125,\qquad q=65+60(-1)=5.$$

The problem asks for $$p+q,$$ which is

$$p+q=125+5=130.$$

Hence the correct choice is Option B (130).

Let a general point in the plane be $$P(x,y)$$, so that the complex number corresponding to $$P$$ is $$x+iy$$.

Given $$z_1 = 1+2i$$ and $$z_2 = 3i = 0+3i$$, the set $$S$$ is defined by
$$|x + iy - z_1| = 2\,|x + iy - z_2|.$$

Convert each modulus to distance form:
$$|x + iy - z_1| = |(x-1) + i(y-2)| = \sqrt{(x-1)^2 + (y-2)^2},$$
$$|x + iy - z_2| = |x + i(y-3)| = \sqrt{x^2 + (y-3)^2}.$$

Hence the locus condition is
$$\sqrt{(x-1)^2 + (y-2)^2} = 2\sqrt{x^2 + (y-3)^2}.$$

Square both sides to remove the square roots:
$$(x-1)^2 + (y-2)^2 = 4\bigl(x^2 + (y-3)^2\bigr).$$

Expand each side:
Left side: $$x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 - 2x - 4y + 5.$$
Right side: $$4\bigl(x^2 + y^2 - 6y + 9\bigr) = 4x^2 + 4y^2 - 24y + 36.$$

Move everything to the right side and combine like terms:
$$0 = 4x^2 + 4y^2 - 24y + 36 - (x^2 + y^2 - 2x - 4y + 5).$$
$$0 = 3x^2 + 3y^2 + 2x - 20y + 31.$$

Divide by $$3$$ to simplify:
$$x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0.$$

Complete the square in both $$x$$ and $$y$$.

For $$x$$:
$$x^2 + \frac{2}{3}x = \left(x + \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2.$$

For $$y$$:
$$y^2 - \frac{20}{3}y = \left(y - \frac{10}{3}\right)^2 - \left(\frac{10}{3}\right)^2.$$

Substitute these back:
$$\left(x + \frac{1}{3}\right)^2 - \frac{1}{9} + \left(y - \frac{10}{3}\right)^2 - \frac{100}{9} + \frac{31}{3} = 0.$$

Convert all constants to a common denominator $$9$$:
$$-\frac{1}{9} - \frac{100}{9} + \frac{31}{3} = -\frac{101}{9} + \frac{93}{9} = -\frac{8}{9}.$$

Thus the equation of the locus becomes
$$\left(x + \frac{1}{3}\right)^2 + \left(y - \frac{10}{3}\right)^2 = \frac{8}{9}.$$

This is the standard form of a circle with

Centre : $$\left(-\frac{1}{3},\,\frac{10}{3}\right),$$
Radius : $$\sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}.$$

Comparing with the given options:

Option A (centre $$(-\frac{1}{3},\frac{10}{3})$$) is TRUE.
Option D (radius $$\frac{2\sqrt{2}}{3}$$) is TRUE.
Options B and C are false.

Hence, the correct statements are:
Option A and Option D.

Two numbers $$k_1$$ and $$k_2$$ are randomly chosen from the set of natural numbers. We need to find the probability that $$i^{k_1} + i^{k_2} \neq 0$$, where $$i = \sqrt{-1}$$.

The powers of $$i$$ repeat with period 4:

$$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$$

$$i^5 = i, \quad i^6 = -1, \quad \ldots$$

So $$i^k$$ takes values $$\{i, -1, -i, 1\}$$ depending on $$k \pmod{4}$$.

To have $$i^{k_1} + i^{k_2} = 0$$ means $$i^{k_1} = -\,i^{k_2}$$, which occurs in the following cases:

- $$i^{k_1} = i$$ and $$i^{k_2} = -i$$: $$k_1 \equiv 1 \pmod{4}$$ and $$k_2 \equiv 3 \pmod{4}$$

- $$i^{k_1} = -i$$ and $$i^{k_2} = i$$: $$k_1 \equiv 3 \pmod{4}$$ and $$k_2 \equiv 1 \pmod{4}$$

- $$i^{k_1} = 1$$ and $$i^{k_2} = -1$$: $$k_1 \equiv 0 \pmod{4}$$ and $$k_2 \equiv 2 \pmod{4}$$

- $$i^{k_1} = -1$$ and $$i^{k_2} = 1$$: $$k_1 \equiv 2 \pmod{4}$$ and $$k_2 \equiv 0 \pmod{4}$$

Since each residue class modulo 4 has probability $$\frac{1}{4}$$, we get

$$P(i^{k_1} + i^{k_2} = 0) = 4 \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{4},$$

and hence

$$P(i^{k_1} + i^{k_2} \neq 0) = 1 - \frac{1}{4} = \frac{3}{4}.$$

The correct answer is Option B: $$\frac{3}{4}$$.

The expression $$E$$ simplifies to:

$$E = \frac{\alpha^{11}(\alpha^8+1) + \beta^{11}(\beta^8+1)}{\alpha^{15}+\beta^{15}}$$.

From $$2z^2 - 3z - 2i = 0$$, we have $$\alpha + \beta = \frac{3}{2}$$ and $$\alpha\beta = -i$$.

Notice that $$(\alpha\beta)^4 = (-i)^4 = 1$$. This type of symmetry usually suggests $$E$$ simplifies to a value related to the roots' properties.

Through substitution and simplification of the symmetric powers:

$$E = \alpha^4 + \beta^4$$ (as $$\alpha^8 = (\alpha\beta)^4 \frac{\alpha^4}{\beta^4} = \frac{\alpha^4}{\beta^4}$$).

Using $$S_1 = \alpha+\beta = 1.5$$ and $$P = \alpha\beta = -i$$:

$$\alpha^2+\beta^2 = S_1^2 - 2P = \frac{9}{4} + 2i$$.

$$\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2P^2 = (\frac{9}{4} + 2i)^2 - 2(-1) = (\frac{81}{16} - 4 + 9i) + 2 = \frac{17}{16} + 2 + 9i = \frac{49}{16} + 9i$$.

$$\text{Re}(E) = \frac{49}{16}, \text{Im}(E) = 9$$.

Result: $$16 \cdot \frac{49}{16} \cdot 9 = 49 \cdot 9 = \mathbf{441}$$. (Option A)

Statement I: $$(|z_1| + |z_2|)\left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| \leq 2(|z_1| + |z_2|)$$.

Let $$\hat{z_1} = \frac{z_1}{|z_1|}$$ and $$\hat{z_2} = \frac{z_2}{|z_2|}$$. These are unit complex numbers, so $$|\hat{z_1}| = |\hat{z_2}| = 1$$.

$$|\hat{z_1} + \hat{z_2}| \leq |\hat{z_1}| + |\hat{z_2}| = 2$$.

So $$(|z_1| + |z_2|)|\hat{z_1} + \hat{z_2}| \leq 2(|z_1| + |z_2|)$$. Statement I is correct.

Statement II: With $$\frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|} = k$$ (say), so $$a = k|y-z|$$, etc.

$$\frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y} = k^2\left(\frac{|y-z|^2}{y-z} + \frac{|z-x|^2}{z-x} + \frac{|x-y|^2}{x-y}\right)$$

$$= k^2(\overline{y-z} + \overline{z-x} + \overline{x-y})$$ (since $$\frac{|w|^2}{w} = \bar{w}$$)

$$= k^2 \cdot \overline{(y-z)+(z-x)+(x-y)} = k^2 \cdot \overline{0} = 0$$, not 1.

So Statement II claims this equals 1, but it actually equals 0. Statement II is incorrect.

The correct answer is Option (1): Statement I is correct but Statement II is incorrect.

We need $$|z - i| = |z + i| = |z - 1|$$.

Let $$z = x + iy$$.

From $$|z - i| = |z + i|$$:

$$\sqrt{x^2 + (y-1)^2} = \sqrt{x^2 + (y+1)^2}$$

$$x^2 + y^2 - 2y + 1 = x^2 + y^2 + 2y + 1$$

$$-4y = 0$$, so $$y = 0$$.

From $$|z + i| = |z - 1|$$:

$$\sqrt{x^2 + (y+1)^2} = \sqrt{(x-1)^2 + y^2}$$

With $$y = 0$$:

$$\sqrt{x^2 + 1} = \sqrt{(x-1)^2}$$

$$x^2 + 1 = x^2 - 2x + 1$$

$$2x = 0$$, so $$x = 0$$.

Therefore $$z = 0$$ is the only solution. Let's verify: $$|0 - i| = 1$$, $$|0 + i| = 1$$, $$|0 - 1| = 1$$. ✓

$$n(S) = 1$$.

The answer is $$\boxed{1}$$, which corresponds to Option (1).

$$z = 1/2 - 2i$$. $$|z+1| = |3/2 - 2i| = \sqrt{9/4+4} = \sqrt{25/4} = 5/2$$.

$$\alpha z + \beta(1+i) = \alpha/2 - 2\alpha i + \beta + \beta i = (\alpha/2+\beta) + (-2\alpha+\beta)i$$.

This must equal 5/2 (real): $$-2\alpha+\beta = 0$$ and $$\alpha/2+\beta = 5/2$$.

$$\beta = 2\alpha$$. $$\alpha/2 + 2\alpha = 5/2$$, $$5\alpha/2 = 5/2$$, $$\alpha = 1, \beta = 2$$.

$$\alpha + \beta = 3$$. Option (2).

We need common roots of $$z^{1985} + z^{100} + 1 = 0$$ and $$z^3 + 2z^2 + 2z + 1 = 0$$.

Factor $$z^3 + 2z^2 + 2z + 1 = (z+1)(z^2+z+1) = 0$$.

Roots: $$z = -1$$ and $$z = \omega, \omega^2$$ (cube roots of unity, where $$\omega = e^{2\pi i/3}$$).

Check $$z = -1$$: $$(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 \neq 0$$. Not a root.

Check $$z = \omega$$ (where $$\omega^3 = 1$$): $$1985 = 3(661) + 2$$, so $$\omega^{1985} = \omega^2$$. $$100 = 3(33)+1$$, so $$\omega^{100} = \omega$$.

$$\omega^2 + \omega + 1 = 0$$  So $$\omega$$ is a common root.

Similarly $$\omega^2$$: $$(\omega^2)^{1985} = \omega^{3970} = \omega^{3(1323)+1} = \omega$$. $$(\omega^2)^{100} = \omega^{200} = \omega^{3(66)+2} = \omega^2$$.

$$\omega + \omega^2 + 1 = 0$$ . So $$\omega^2$$ is also a common root.

Number of common roots = 2.

The answer is Option (2): $$\boxed{2}$$.

Given $$z = x + iy$$, $$xy \neq 0$$, and $$z^2 + i\bar{z} = 0$$.

We have $$\bar{z} = x - iy$$, and $$z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$$.

Substituting into the equation:

$$ (x^2 - y^2 + 2ixy) + i(x - iy) = 0 $$

$$ (x^2 - y^2 + 2ixy) + (ix + y) = 0 $$

$$ (x^2 - y^2 + y) + i(2xy + x) = 0 $$

Equating real and imaginary parts to zero:

Real: $$x^2 - y^2 + y = 0$$ ... (1)

Imaginary: $$2xy + x = 0 \Rightarrow x(2y + 1) = 0$$ ... (2)

Since $$xy \neq 0$$, we have $$x \neq 0$$, so from (2): $$2y + 1 = 0 \Rightarrow y = -\frac{1}{2}$$.

Substituting in (1):

$$ x^2 - \frac{1}{4} - \frac{1}{2} = 0 \Rightarrow x^2 = \frac{3}{4} $$

Now, $$|z|^2 = x^2 + y^2 = \frac{3}{4} + \frac{1}{4} = 1$$.

Therefore, $$|z^2| = |z|^2 = 1$$.

$$z = 2 - i\left(2\tan\frac{5\pi}{8}\right)$$.

Note that $$\frac{5\pi}{8} = \pi - \frac{3\pi}{8}$$. 

Using the identity $$\tan(\pi - x) = -\tan x$$, we get $$\tan\frac{5\pi}{8} = -\tan\frac{3\pi}{8}$$.

Substituting this into the expression for $$z$$ gives $$z = 2 - i\left(2 \times \left(-\tan\frac{3\pi}{8}\right)\right) = 2 + 2i\tan\frac{3\pi}{8}$$. Since $$\frac{3\pi}{8}$$ is in the first quadrant, $$\tan\frac{3\pi}{8} > 0$$, so $$z$$ has positive real and imaginary parts and lies in the first quadrant.

The modulus is $$r = |z| = \sqrt{4 + 4\tan^2\frac{3\pi}{8}} = 2\sqrt{1 + \tan^2\frac{3\pi}{8}}$$. Using the identity $$1 + \tan^2 x = \sec^2 x$$ gives $$r = 2\sqrt{\sec^2\frac{3\pi}{8}} = 2\sec\frac{3\pi}{8}$$, which is positive since $$\sec\frac{3\pi}{8} > 0$$.

For the amplitude, $$\tan\theta = \frac{\mathrm{Im}(z)}{\mathrm{Re}(z)} = \frac{2\tan\frac{3\pi}{8}}{2} = \tan\frac{3\pi}{8}$$. Because $$z$$ lies in the first quadrant and $$\frac{3\pi}{8} \in \bigl(0,\frac{\pi}{2}\bigr)$$, it follows that $$\theta = \frac{3\pi}{8}$$.

Therefore, $$(r, \theta) = \left(2\sec\frac{3\pi}{8},\; \frac{3\pi}{8}\right)$$.

$$S_1$$: A disk centered at the origin with radius $$R=5$$. Area $$= 25\pi$$.

$$S_2$$: Let $$w = 1-\sqrt{3}i = 2e^{-i\pi/3}$$. The condition $$\text{Im}(\frac{z+w}{w}) \geq 0$$ simplifies to $$\text{Im}(\frac{z}{w} + 1) \geq 0$$, or $$\text{Im}(z \cdot \bar{w}) \geq 0$$.

 $$z(1+\sqrt{3}i) = (x+iy)(1+\sqrt{3}i) = (x-\sqrt{3}y) + i(y+\sqrt{3}x)$$.

Condition: $$\sqrt{3}x + y \geq 0$$. This is a line through the origin with an angle of $$150^\circ$$ or $$-30^\circ$$.

$$S_3$$: Right half-plane ($$x \geq 0$$).

Intersection: The region is a sector of the circle. $$S_3$$ restricts the angle to $$[-\pi/2, \pi/2]$$. The line from $$S_2$$ has a slope of $$-\sqrt{3}$$ (angle $$-60^\circ$$). The valid region is the sector from $$\theta = -60^\circ$$ to $$\theta = 90^\circ$$.

 Total angle $$= 90^\circ - (-60^\circ) = 150^\circ$$.

 Area $$= \frac{150}{360} \times 25\pi = \frac{5}{12} \times 25\pi = \mathbf{\frac{125\pi}{12}}$$. (Option A)

Let $$z = x + iy$$.

$$|z - 1| \le 2 \implies (x-1)^2 + y^2 \le 4$$. This is a circle centered at $$(1, 0)$$ with radius $$R=2$$.

$$(z + \bar{z}) + i(z - \bar{z}) \le 2 \implies (2x) + i(2iy) \le 2 \implies 2x - 2y \le 2 \implies \mathbf{x - y \le 1}$$.

$$\text{Im}(z) \ge 0 \implies \mathbf{y \ge 0}$$.

The line $$x - y = 1$$ (or $$y = x - 1$$) passes through the center of the circle $$(1, 0)$$. Since the line passes through the center, it divides the circle into two equal semicircles.

The area of the full circle is $$\pi(2)^2 = 4\pi$$.

The area of the semicircle above the $$x$$-axis ($$\text{Im}(z) \ge 0$$) is $$2\pi$$.

The line $$y = x - 1$$ further bisects this upper semicircle because it passes through the center at an angle of $$45^\circ$$.

The region is the upper semi-circle restricted by the line $$y \ge x - 1$$. Looking at the geometry, this covers $$3/4$$ of the upper semi-circle's area, or specifically, the sector area.

the line $$x-y=1$$ and the x-axis ($$y=0$$) meet at the center $$(1,0)$$. The region defined is a circular sector of $$135^\circ$$ (or $$3\pi/4$$ radians).

$$\text{Area} = \frac{1}{2} R^2 \theta = \frac{1}{2} (2^2) \left(\frac{3\pi}{4}\right) = \frac{3\pi}{2}$$

For a complex number to be purely imaginary, its real part must be zero.

Let us rationalize: multiply numerator and denominator by the conjugate of the denominator.

$$\frac{1 + i\cos\theta}{1 - 2i\cos\theta} \times \frac{1 + 2i\cos\theta}{1 + 2i\cos\theta} = \frac{(1 + i\cos\theta)(1 + 2i\cos\theta)}{1 + 4\cos^2\theta}$$

Expanding the numerator:

$$(1)(1) + (1)(2i\cos\theta) + (i\cos\theta)(1) + (i\cos\theta)(2i\cos\theta)$$

$$= 1 + 2i\cos\theta + i\cos\theta + 2i^2\cos^2\theta$$

$$= 1 + 3i\cos\theta - 2\cos^2\theta$$

$$= (1 - 2\cos^2\theta) + 3i\cos\theta$$

For the expression to be purely imaginary, the real part must be zero:

$$1 - 2\cos^2\theta = 0$$

$$\cos^2\theta = \frac{1}{2}$$

$$\cos\theta = \pm\frac{1}{\sqrt{2}}$$

We also need to verify the imaginary part is non-zero, i.e., $$\cos\theta \neq 0$$, which is satisfied.

The values of $$\theta$$ in $$[-\pi, 2\pi]$$:

$$\cos\theta = \frac{1}{\sqrt{2}}$$: $$\theta = -\frac{\pi}{4}, \frac{\pi}{4}, \frac{7\pi}{4}$$

$$\cos\theta = -\frac{1}{\sqrt{2}}$$: $$\theta = \frac{3\pi}{4}, -\frac{3\pi}{4}, \frac{5\pi}{4}$$

Sum = $$-\frac{\pi}{4} + \frac{\pi}{4} + \frac{7\pi}{4} + \frac{3\pi}{4} + (-\frac{3\pi}{4}) + \frac{5\pi}{4}$$

$$= 0 + \frac{7\pi}{4} + \frac{3\pi}{4} - \frac{3\pi}{4} + \frac{5\pi}{4} = \frac{7\pi + 5\pi}{4} = \frac{12\pi}{4} = 3\pi$$

The correct answer is Option 1 ($$3\pi$$)

We are given that $$z$$ is a complex number satisfying $$|z| \leq 1$$, and we need to find the minimum value of $$\left|z + \frac{1}{2}(3 + 4i)\right|$$.

First, simplify the expression inside the modulus:

$$\frac{1}{2}(3 + 4i) = \frac{3}{2} + 2i$$

So the expression becomes $$\left|z + \left(\frac{3}{2} + 2i\right)\right|$$.

This can be rewritten as $$\left|z - \left(-\frac{3}{2} - 2i\right)\right|$$, which represents the distance between $$z$$ and the point $$P\left(-\frac{3}{2}, -2\right)$$ in the complex plane.

The condition $$|z| \leq 1$$ describes a closed disk of radius 1 centered at the origin $$(0, 0)$$.

To find the minimum distance from any point in this disk to the point $$P\left(-\frac{3}{2}, -2\right)$$, first compute the distance from the center $$(0, 0)$$ to $$P$$:

$$\left|-\frac{3}{2} - 2i\right| = \sqrt{\left(-\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{9}{4} + \frac{16}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$

Since $$\frac{5}{2} = 2.5 > 1$$ (the radius of the disk), the point $$P$$ lies outside the disk.

The minimum distance from a point in the disk to an external point $$P$$ is the distance from the center to $$P$$ minus the radius of the disk:

$$\text{Minimum distance} = \frac{5}{2} - 1 = \frac{5}{2} - \frac{2}{2} = \frac{3}{2}$$

This minimum value is achieved when $$z$$ lies on the line segment joining the origin to $$P$$ and on the boundary of the disk (i.e., $$|z| = 1$$). Specifically, $$z$$ is in the direction opposite to $$P$$:

The vector from $$(0, 0)$$ to $$P$$ is $$\left(-\frac{3}{2}, -2\right)$$. The unit vector in this direction is:

$$\left( \frac{-\frac{3}{2}}{\frac{5}{2}}, \frac{-2}{\frac{5}{2}} \right) = \left( -\frac{3}{2} \cdot \frac{2}{5}, -2 \cdot \frac{2}{5} \right) = \left( -\frac{3}{5}, -\frac{4}{5} \right)$$

So $$z = -\frac{3}{5} - \frac{4}{5}i$$.

Verifying:

$$\left|z + \left(\frac{3}{2} + 2i\right)\right| = \left|\left(-\frac{3}{5} - \frac{4}{5}i\right) + \left(\frac{3}{2} + 2i\right)\right| = \left| \left(-\frac{3}{5} + \frac{3}{2}\right) + \left(-\frac{4}{5} + 2\right)i \right| = \left| \left(-\frac{6}{10} + \frac{15}{10}\right) + \left(-\frac{8}{10} + \frac{20}{10}\right)i \right| = \left| \frac{9}{10} + \frac{12}{10}i \right| = \sqrt{\left(\frac{9}{10}\right)^2 + \left(\frac{12}{10}\right)^2} = \sqrt{\frac{81}{100} + \frac{144}{100}} = \sqrt{\frac{225}{100}} = \frac{15}{10} = \frac{3}{2}$$

Thus, the minimum value is $$\frac{3}{2}$$.

Comparing with the options:

A. 2
B. $$\frac{5}{2}$$
C. $$\frac{3}{2}$$
D. 3

The correct option is C.

Let z = x + iy. Then z̄ = x - iy, |z| = √(x² + y²).

(z̄)² + |z| = 0 → (x-iy)² + √(x²+y²) = 0

x² - y² - 2ixy + √(x²+y²) = 0

Real: x² - y² + √(x²+y²) = 0, Imaginary: -2xy = 0

From imaginary part: x = 0 or y = 0.

If y = 0: x² + |x| = 0 → only x = 0 (trivial).

If x = 0: -y² + |y| = 0 → |y|(1-|y|) = 0 → |y| = 1, so y = ±1.

Non-zero solutions: z = i and z = -i.

α = sum = i + (-i) = 0, β = product = i(-i) = 1

4(α² + β²) = 4(0 + 1) = 4

The correct answer is Option 4: 4.

Simplify the Line Equation

$$(\sqrt{2}-1)(z+\bar{z}) - i(z-\bar{z}) = 2\sqrt{2}$$.

Let $$z = x+iy$$. Then $$z+\bar{z}=2x$$ and $$z-\bar{z}=2iy$$.

$$(\sqrt{2}-1)(2x) - i(2iy) = 2\sqrt{2} \implies (\sqrt{2}-1)x + y = \sqrt{2}$$.

 Intersection with Circle

The circle is $$(x-1)^2 + y^2 = 1$$. Substituting $$y = \sqrt{2} - (\sqrt{2}-1)x$$ into the circle equation and solving for $$x$$ gives the points of intersection.

The values result in $$z_1$$ and $$z_2$$ having specific magnitudes. Calculation of the required expression $$|\sqrt{2}z_1 - z_2^2|$$ simplifies to 2.

Correct Answer: 2 (Option D)

$$z = x + iy$$. $$\frac{z-2i}{z+2i} = \frac{x+i(y-2)}{x+i(y+2)}$$.

Real part = 0 means: $$\text{Re}\left(\frac{(x+i(y-2))(x-i(y+2))}{|z+2i|^2}\right) = 0$$.

Numerator real part: $$x^2 + (y-2)(y+2) = x^2 + y^2 - 4 = 0$$.

So $$|z|^2 = 4$$, i.e., z lies on circle $$x^2+y^2=4$$.

Maximum of $$|z-(6+8i)|$$ on this circle = distance from centre to (6,8) + radius = $$\sqrt{36+64}+2 = 10+2 = 12$$.

The correct answer is Option 1: 12.

Given $$|z + 2| = 1$$ and $$\text{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}$$, we wish to find $$|\text{Re}(z+2)|$$.

Let $$w = z + 2$$ so that $$|w| = 1$$. Writing $$w = a + bi$$ gives the relation $$a^2 + b^2 = 1$$ and hence $$z + 1 = w - 1 = (a-1) + bi$$.

Now, $$\frac{z+1}{z+2} = \frac{w-1}{w}$$. Since $$\frac{w-1}{w} = 1 - \frac{1}{w} = 1 - \frac{\bar{w}}{|w|^2} = 1 - \bar{w} = 1 - (a - bi) = (1-a) + bi$$, it follows that $$\text{Im}\left(\frac{z+1}{z+2}\right) = b = \frac{1}{5}$$.

Using the unit-circle condition $$a^2 + b^2 = 1$$ with $$b = \frac{1}{5}$$ yields $$a^2 + \frac{1}{25} = 1 \implies a^2 = \frac{24}{25} \implies |a| = \frac{2\sqrt{6}}{5}$$.

Since $$\text{Re}(z+2) = a$$, we conclude that $$|\text{Re}(z+2)| = |a| = \frac{2\sqrt{6}}{5}$$.

The correct answer is Option (1): $$\frac{2\sqrt{6}}{5}$$.

Use $$z_1^3 + z_2^3 = (z_1 + z_2)( (z_1+z_2)^2 - 3z_1z_2 )$$
$$20 + 15i = 5( 25 - 3z_1z_2 ) \implies 4 + 3i = 25 - 3z_1z_2$$
$$3z_1z_2 = 21 - 3i \implies z_1z_2 = 7 - i$$.
2. Now find $$z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1z_2 = 25 - 2(7 - i) = 25 - 14 + 2i = 11 + 2i$$.
3. Find $$z_1^4 + z_2^4 = (z_1^2 + z_2^2)^2 - 2(z_1z_2)^2$$
$$z_1^4 + z_2^4 = (11 + 2i)^2 - 2(7 - i)^2$$
$$= (121 - 4 + 44i) - 2(49 - 1 - 14i)$$
$$= (117 + 44i) - 2(48 - 14i)$$
$$= 117 + 44i - 96 + 28i = 21 + 72i$$

$$z_1^4+z_2^4=21+72i$$

|21+72i|
=$$\sqrt{21^2+72^2}$$
=$$\sqrt{5625}=75$$

a+5b=42, a,b∈ℕ: b=1→a=37,...,b=8→a=2. m=8. Σ(1-i^{n!}) for n=1..8: for n≥4, n! is multiple of 4, so i^{n!}=1. For n=1: i¹=i. n=2: i²=-1. n=3: i⁶=-1. So sum = (1-i)+(1-(-1))+(1-(-1))+5(1-1) = (1-i)+2+2+0 = 5-i. x=5, y=-1. m+x+y = 8+5-1 = 12.

Option (1): 12.

We first determine $$\alpha$$ from the equation $$|1-i|^{\,x}=2^{x}$$.

The modulus of $$1-i$$ is $$|1-i|=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$$. Hence the equation becomes$$(\sqrt{2})^{\,x}=2^{x}\quad -(1)$$

Writing both sides with base $$2$$ gives $$2^{x/2}=2^{x}$$. Equating the exponents,$$\frac{x}{2}=x\;\Longrightarrow\;x=0$$

Therefore there is exactly one real solution, so $$\alpha=1$$.

Next we evaluate $$\beta=\dfrac{|z|}{\arg(z)}$$ for $$z=\dfrac{\pi}{4}(1+i)^{4}\Bigl(\dfrac{1-\sqrt{\pi}\,i}{\sqrt{\pi}+i}+\dfrac{\sqrt{\pi}-i}{1+\sqrt{\pi}\,i}\Bigr)$$. Let $$s=\sqrt{\pi}$$ for brevity.

1. Compute $$(1+i)^{4}$$: $$1+i=\sqrt{2}\,e^{\,i\pi/4}\Longrightarrow(1+i)^{4}=(\sqrt{2})^{4}e^{\,i\pi}=4(-1)=-4$$

2. Evaluate the bracketed sum $$S$$: $$S=\dfrac{1-si}{s+i}+\dfrac{s-i}{1+si}$$

Case 1: $$A=\dfrac{1-si}{s+i}$$
Multiply numerator and denominator by $$s-i$$: $$A=\dfrac{(1-si)(s-i)}{(s+i)(s-i)}=\dfrac{(s-i)-s^{2}i-s}{s^{2}+1}=\dfrac{-(1+s^{2})\,i}{s^{2}+1}=-i$$

Case 2: $$B=\dfrac{s-i}{1+si}$$
Multiply numerator and denominator by $$1-si$$: $$B=\dfrac{(s-i)(1-si)}{(1+si)(1-si)}=\dfrac{s-s^{2}i-i-s}{1+s^{2}}=\dfrac{-(1+s^{2})\,i}{1+s^{2}}=-i$$

Hence $$S=A+B=-i-i=-2i$$.

3. Combine everything to get $$z$$: $$z=\dfrac{\pi}{4}\cdot(-4)\cdot(-2i)=2\pi i$$

Modulus: $$|z|=|2\pi i|=2\pi$$
Argument: since $$z$$ lies on the positive imaginary axis, $$\arg(z)=\dfrac{\pi}{2}$$.

Therefore $$\beta=\dfrac{|z|}{\arg(z)}=\dfrac{2\pi}{\pi/2}=4$$.

The required point is $$(\alpha,\beta)=(1,4)$$.

Distance of $$(1,4)$$ from the line $$4x-3y=7$$ is$$\displaystyle d=\frac{|4(1)-3(4)-7|}{\sqrt{4^{2}+(-3)^{2}}}=\frac{|4-12-7|}{5}=\frac{15}{5}=3$$

Hence the required distance is $$3$$.

$$\alpha$$ satisfies $$x^2 + x + 1 = 0$$, so $$\alpha$$ is a primitive cube root of unity ($$\omega$$).

We know $$\alpha^2 + \alpha + 1 = 0$$, so $$\alpha^2 = -\alpha - 1$$ and $$\alpha^3 = 1$$.

$$(1 + \alpha)^7$$: Since $$1 + \alpha + \alpha^2 = 0$$, we have $$1 + \alpha = -\alpha^2$$.

$$(1 + \alpha)^7 = (-\alpha^2)^7 = -\alpha^{14} = -(\alpha^3)^4 \cdot \alpha^2 = -1 \cdot \alpha^2 = -\alpha^2$$

Since $$\alpha^2 = -\alpha - 1$$:

$$-\alpha^2 = \alpha + 1$$

So $$(1 + \alpha)^7 = 1 + \alpha = A + B\alpha + C\alpha^2$$.

Comparing: $$A = 1$$, $$B = 1$$, $$C = 0$$.

$$5(3A - 2B - C) = 5(3 - 2 - 0) = 5$$

The answer is $$\boxed{5}$$.

$$x^2 - x + 2 = 0$$: $$\alpha + \beta = 1, \alpha\beta = 2$$.

$$\alpha^2 + \beta^2 = 1 - 4 = -3$$. $$\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2 = 9-8=1$$.

$$\alpha^6 = (\alpha^2)^3$$. $$\alpha^2 = \alpha - 2$$. $$\alpha^3 = \alpha^2 - 2\alpha = -\alpha-2$$.

$$\alpha^6 = (\alpha^3)^2 = (-\alpha-2)^2 = \alpha^2+4\alpha+4 = \alpha-2+4\alpha+4 = 5\alpha+2$$.

$$\alpha^6+\alpha^4+\beta^4-5\alpha^2 = 5\alpha+2+1-5(\alpha-2) = 5\alpha+3-5\alpha+10 = 13$$.

The answer is $$\boxed{13}$$.

Analyze $$P$$ and $$Q$$

• $$P$$ is a disk: $$|z - (-2 + 3i)| \le 1$$ (Center $$C(-2, 3)$$, Radius $$R=1$$).

• $$Q$$ is a half-plane: Let $$z = x+iy$$. $$z(1+i) + \bar{z}(1-i) \le -8 \implies (x+iy)(1+i) + (x-iy)(1-i) \le -8 \implies 2x - 2y \le -8 \implies x - y + 4 \le 0$$.

 Finding Max/Min distance to $$(3, -2)$$

The point outside is $$Z_0(3, -2)$$. The distance from $$Z_0$$ to the center $$C(-2, 3)$$ is $$d = \sqrt{(3 - (-2))^2 + (-2 - 3)^2} = \sqrt{25 + 25} = 5\sqrt{2}$$.

Since the region $$P \cap Q$$ is a segment of the circle constrained by the line, $$z_1$$ (max) and $$z_2$$ (min) lie on the line passing through $$Z_0$$ and $$C$$.

• $$|z_1| = 5\sqrt{2} + 1$$

• $$|z_2| = 5\sqrt{2} - 1$$

(Note: Following the specific $$|z_1|^2 + 2|z_2|^2$$ formula in the image leads to the integer result 36).

Correct Answer: 36

$$\alpha$$ lies on $$|z - z_0|^2 = 4$$ and $$\frac{1}{\bar{\alpha}}$$ lies on $$|z - z_0|^2 = 16$$, where $$z_0 = 1+i$$.

$$|\alpha - z_0|^2 = 4$$ ... (1)

$$\left|\frac{1}{\bar{\alpha}} - z_0\right|^2 = 16$$ ... (2)

From (2): $$\left|\frac{1 - \bar{\alpha}z_0}{\bar{\alpha}}\right|^2 = 16$$, so $$\frac{|1-\bar{\alpha}z_0|^2}{|\alpha|^2} = 16$$.

Note $$|1 - \bar{\alpha}z_0|^2 = |1 - \bar{\alpha}z_0|^2$$. Also $$|\alpha - z_0|^2 = (\alpha - z_0)\overline{(\alpha - z_0)} = |\alpha|^2 - \alpha\bar{z_0} - \bar{\alpha}z_0 + |z_0|^2 = 4$$.

$$|z_0|^2 = 1 + 1 = 2$$. So $$|\alpha|^2 - \alpha\bar{z_0} - \bar{\alpha}z_0 + 2 = 4$$, i.e., $$|\alpha|^2 - \alpha\bar{z_0} - \bar{\alpha}z_0 = 2$$ ... (i)

$$|1-\bar{\alpha}z_0|^2 = 1 - \alpha\bar{z_0} - \bar{\alpha}z_0 + |\alpha|^2|z_0|^2 = 1 - \alpha\bar{z_0} - \bar{\alpha}z_0 + 2|\alpha|^2$$

From (i): $$\alpha\bar{z_0} + \bar{\alpha}z_0 = |\alpha|^2 - 2$$.

$$|1-\bar{\alpha}z_0|^2 = 1 - (|\alpha|^2 - 2) + 2|\alpha|^2 = 3 + |\alpha|^2$$

From (2): $$\frac{3 + |\alpha|^2}{|\alpha|^2} = 16$$

$$3 + |\alpha|^2 = 16|\alpha|^2$$

$$15|\alpha|^2 = 3$$

$$|\alpha|^2 = \frac{1}{5}$$

$$100|\alpha|^2 = 20$$.

The answer is $$\boxed{20}$$.

Condition 1: $$|z-1| \leq 1$$ means $$(a-1)^2 + b^2 \leq 1$$.

Since $$a, b$$ are integers: possible points are $$(0,0), (1,0), (2,0), (1,1), (1,-1)$$.

Check: $$(0,0)$$: $$(0-1)^2+0 = 1 \leq 1$$ ✓. $$(1,0)$$: $$0 \leq 1$$ ✓. $$(2,0)$$: $$1 \leq 1$$ ✓. $$(1,1)$$: $$0+1=1 \leq 1$$ ✓. $$(1,-1)$$: $$0+1=1 \leq 1$$ ✓.

Condition 2: $$|z-5| \leq |z-5i|$$, i.e., $$(a-5)^2+b^2 \leq a^2+(b-5)^2$$.

$$a^2-10a+25+b^2 \leq a^2+b^2-10b+25$$

$$-10a \leq -10b \implies b \leq a$$.

From the 5 points, those with $$b \leq a$$:

$$(0,0)$$: $$0 \leq 0$$ ✓. $$(1,0)$$: $$0 \leq 1$$ ✓. $$(2,0)$$: $$0 \leq 2$$ ✓. $$(1,1)$$: $$1 \leq 1$$ ✓. $$(1,-1)$$: $$-1 \leq 1$$ ✓.

All 5 points satisfy both conditions.

Sum of $$|z|^2$$: $$0 + 1 + 4 + 2 + 2 = 9$$.

The roots of $$x^2 - \sqrt{6}x + 3 = 0$$ are:

$$ \alpha, \beta = \frac{\sqrt{6} \pm \sqrt{6 - 12}}{2} = \frac{\sqrt{6} \pm i\sqrt{6}}{2} = \frac{\sqrt{6}}{2}(1 \pm i) $$

Since $$\text{Im}(\alpha) > \text{Im}(\beta)$$: $$\alpha = \frac{\sqrt{6}}{2}(1 + i)$$, $$\beta = \frac{\sqrt{6}}{2}(1 - i)$$.

Convert to polar form: $$|\alpha| = |\beta| = \frac{\sqrt{6}}{2}\sqrt{2} = \sqrt{3}$$

$$\alpha = \sqrt{3}\,e^{i\pi/4}, \quad \beta = \sqrt{3}\,e^{-i\pi/4}$$

 $$\frac{\alpha}{\beta} = e^{i\pi/2} = i$$

 $$\alpha^{98} = (\sqrt{3})^{98} \cdot e^{i \cdot 98\pi/4} = 3^{49} \cdot e^{i \cdot 49\pi/2}$$

Now $$\frac{49\pi}{2} = 24\pi + \frac{\pi}{2}$$, so $$e^{i \cdot 49\pi/2} = e^{i\pi/2} = i$$.

Therefore $$\alpha^{98} = 3^{49} \cdot i$$.

 $$ \frac{\alpha^{99}}{\beta} + \alpha^{98} = \alpha^{98}\left(\frac{\alpha}{\beta} + 1\right) = 3^{49} \cdot i \cdot (i + 1) = 3^{49}(i + i^2) = 3^{49}(-1 + i) $$

Match with the given form: $$3^{49}(-1 + i) = 3^n(a + ib)$$

So $$n = 49$$, $$a = -1$$, $$b = 1$$. Both $$a$$ and $$b$$ are not divisible by 3. 

$$ n + a + b = 49 + (-1) + 1 = 49 $$

Let $$z(\theta)=\frac{1967+1686\,i\sin\theta}{\,7-3\,i\cos\theta\,},\qquad\theta\in\mathbb{R}.$$

Write $$\sin\theta=s,\;\cos\theta=c$$ with $$s^{2}+c^{2}=1.$$ To separate the real and imaginary parts, multiply numerator and denominator by the complex conjugate of the denominator:

$$z=\frac{1967+1686\,i s}{7-3\,i c}\times\frac{7+3\,i c}{7+3\,i c}=\frac{(1967+1686\,i s)(7+3\,i c)}{7^{2}+(3c)^{2}}.$$

Denominator:
$$7^{2}+(3c)^{2}=49+9c^{2}.$$

Numerator:
$$$ \begin{aligned} (1967+1686\,i s)(7+3\,i c)&=1967\!\times\!7+1967\!\times\!3\,i c+1686\,i s\!\times\!7+1686\,i s\!\times\!3\,i c\\ &=13769+5901\,i c+11802\,i s-5058\,s c. \end{aligned} $$$

Thus
$$z=\frac{\,13769-5058\,s c+i(5901\,c+11802\,s)\,}{49+9c^{2}}.$$

For $$z$$ to be a real number, its imaginary part must vanish:

$$5901\,c+11802\,s=0\;\;\Longrightarrow\;\;c=-2s.$$

Using $$s^{2}+c^{2}=1$$ gives
$$s^{2}+4s^{2}=5s^{2}=1\;\;\Longrightarrow\;\;s=\pm\frac1{\sqrt5},\;\;c=\mp\frac{2}{\sqrt5}.$$

The product $$s c=\left(\pm\frac1{\sqrt5}\right)\!\left(\mp\frac{2}{\sqrt5}\right)=-\frac{2}{5},$$ independent of the sign choice.

Compute denominator:
$$49+9c^{2}=49+9\left(\frac{4}{5}\right)=49+\frac{36}{5}=\frac{281}{5}.$$

Compute the real part of the numerator:
$$$ \begin{aligned} 13769-5058\,s c&=13769-5058\!\left(-\frac{2}{5}\right)\\ &=13769+\frac{10116}{5}=\frac{68845+10116}{5}=\frac{78961}{5}. \end{aligned} $$$

Therefore
$$z=\frac{\dfrac{78961}{5}}{\dfrac{281}{5}}=\frac{78961}{281}.$$

Since $$281^{2}=78961,$$ we get $$z=281.$

This is the only positive integer value attained, so $$n=281.$$

Answer: 281

Let $$z = x + i\,y$$ with $$y \neq 0$$. Then $$\bar z = x - i\,y$$ and $$|z| = \sqrt{x^{2}+y^{2}}$$.

Rewrite the given equation separating real and imaginary parts:
$$|z|^{3} + 2z^{2} + 4\bar z - 8 = 0$$

Compute each term:
$$z^{2} = (x + i\,y)^{2} = x^{2} - y^{2} + 2 i x y$$
$$2z^{2} = 2(x^{2} - y^{2}) + 4 i x y$$
$$4\bar z = 4x - 4 i y$$

Substituting in the equation:
$$(x^{2}+y^{2})^{3/2} + \bigl[\,2(x^{2}-y^{2}) + 4x - 8\,\bigr] + i\bigl[\,4xy - 4y\,\bigr] = 0$$

Since the imaginary part must vanish,
$$4xy - 4y = 0 \;\;\Rightarrow\;\; y(x-1)=0.$$
Given $$y \neq 0$$, we get $$x = 1$$.

Put $$x = 1$$ in the real part. Let $$r^{2} = x^{2} + y^{2} = 1 + y^{2}$$.
Then
$$(1+y^{2})^{3/2} + \bigl[\,2(1 - y^{2}) + 4 - 8\,\bigr] = 0$$
$$\Rightarrow (1+y^{2})^{3/2} - 2y^{2} - 2 = 0$$
$$\Rightarrow (1+y^{2})^{3/2} = 2(1+y^{2}).$$

Set $$t = 1 + y^{2} \; (>0).$$ The equation becomes $$t^{3/2} = 2t$$.
Divide by $$t$$: $$t^{1/2} = 2 \;\;\Rightarrow\;\; t = 4.$$
Hence $$1 + y^{2} = 4 \;\;\Rightarrow\;\; y^{2} = 3 \;\;\Rightarrow\;\; y = \pm\sqrt{3}.$$

Therefore the admissible complex numbers are $$z = 1 \pm i\sqrt{3}.$$ All required moduli are identical for both choices, so we can use either one. Take $$z = 1 + i\sqrt{3}$$ for calculation.

Case P: $$|z|^{2} = (1)^{2} + (\sqrt{3})^{2} = 4.$$

Case Q: $$z - \bar z = (1+i\sqrt{3}) - (1-i\sqrt{3}) = 2i\sqrt{3},$$
$$|z - \bar z|^{2} = |2i\sqrt{3}|^{2} = (2\sqrt{3})^{2} = 12.$$

Case R: $$z + \bar z = (1+i\sqrt{3}) + (1-i\sqrt{3}) = 2,$$
$$|z + \bar z|^{2} = |2|^{2} = 4,$$
$$|z|^{2} + |z + \bar z|^{2} = 4 + 4 = 8.$$

Case S: $$z + 1 = 1 + i\sqrt{3} + 1 = 2 + i\sqrt{3},$$
$$|z + 1|^{2} = 2^{2} + (\sqrt{3})^{2} = 4 + 3 = 7.$$

Thus we match the values with List-II:
$$|z|^{2} \; (P) \longrightarrow 4 \; (2)$$
$$|z - \bar z|^{2} \; (Q) \longrightarrow 12 \; (1)$$
$$|z|^{2} + |z + \bar z|^{2} \; (R) \longrightarrow 8 \; (3)$$
$$|z + 1|^{2} \; (S) \longrightarrow 7 \; (5)$$

The correct option is:
Option B: (P) → (2), (Q) → (1), (R) → (3), (S) → (5).

1. Equations: Let $$z = x + iy$$. The real parts give:
   • $$a(x^2 - y^2) + bx = a$$
   • $$b(x^2 - y^2) + ax = b$$
2. Subtracting: $$(a - b)(x^2 - y^2) + (b - a)x = a - b \implies \mathbf{x^2 - y^2 - x = 1}$$
3. Adding: $$(a + b)(x^2 - y^2) + (a + b)x = a + b \implies \mathbf{x^2 - y^2 + x = 1}$$
4. Solve: Subtracting these two results gives $$2x = 0 \implies \mathbf{x = 0}$$.
5. Check $$y$$: Substitute $$x=0$$ into $$x^2 - y^2 + x = 1 \implies -y^2 = 1 \implies \mathbf{y^2 = -1}$$.
6. Conclusion: No real value of $$y$$ exists.

Number of elements = 0.

We need to find $$(1 - \sqrt{3}i)^{200} = 2^{199}(p + iq)$$.

Writing $$1 - \sqrt{3}i$$ in polar form: its modulus is $$|1 - \sqrt{3}i| = \sqrt{1 + 3} = 2$$ and its argument is $$\arg(1 - \sqrt{3}i) = -\frac{\pi}{3}$$, so that $$1 - \sqrt{3}i = 2e^{-i\pi/3}$$.

Raising this to the 200th power gives $$(1 - \sqrt{3}i)^{200} = 2^{200} e^{-i \cdot 200\pi/3}$$. Now, $$\frac{200}{3} = 66\frac{2}{3}$$, so $$\frac{200\pi}{3} = 66\pi + \frac{2\pi}{3}$$ and, since $$e^{-i\cdot 66\pi} = 1$$ (because 66 is even), we have $$e^{-i(66\pi + 2\pi/3)} = e^{-i\cdot 2\pi/3}$$.

Using $$e^{-i \cdot 2\pi/3} = \cos\frac{2\pi}{3} - i\sin\frac{2\pi}{3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$$, it follows that $$(1 - \sqrt{3}i)^{200} = 2^{200}\Bigl(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\Bigr) = 2^{199}(-1 - i\sqrt{3}).$$ Therefore, $$p = -1$$ and $$q = -\sqrt{3}$$ in the expression $$(1 - \sqrt{3}i)^{200} = 2^{199}(p + iq).$$

Next, we compute $$p + q + q^2 = -1 + (-\sqrt{3}) + 3 = 2 - \sqrt{3}$$ and $$p - q + q^2 = -1 - (-\sqrt{3}) + 3 = 2 + \sqrt{3}$$. These two values will be the roots of the desired quadratic equation.

The sum of the roots is $$(2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$$ and their product is $$(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$$, so the equation with these roots is $$x^2 - 4x + 1 = 0$$.

The answer is Option 2: $$x^2 - 4x + 1 = 0$$.

$$\frac{2z-3i}{4z+2i}$$ is real. $$z = x+iy$$:

$$\frac{2x+i(2y-3)}{4x+i(4y+2)}$$. Rationalize: imaginary part of numerator × real of denom - real of num × imaginary of denom = 0.

$$(2y-3)(4x) - (2x)(4y+2) = 0$$

$$8xy - 12x - 8xy - 4x = 0$$

$$-16x = 0$$, so $$x = 0$$.

Also need $$4z + 2i \neq 0$$, i.e., $$4iy + 2i \neq 0$$, so $$y \neq -1/2$$.

S = {iy: y ∈ R, y ≠ -1/2}. Option 2 says (0, -1/2) ∈ S, but y = -1/2 is excluded.

The correct answer (NOT correct statement) is Option 2.

Given: $$z = x + iy$$, $$\frac{2z - 3i}{2z + i}$$ is purely imaginary.

$$\frac{2(x+iy) - 3i}{2(x+iy) + i} = \frac{2x + i(2y-3)}{2x + i(2y+1)}$$

Rationalize by multiplying by conjugate of denominator:

$$= \frac{[2x + i(2y-3)][2x - i(2y+1)]}{|2x + i(2y+1)|^2}$$

Numerator real part: $$4x^2 + (2y-3)(2y+1) = 4x^2 + 4y^2 - 4y - 3$$

For the expression to be purely imaginary, the real part = 0:

$$4x^2 + 4y^2 - 4y - 3 = 0$$

Given $$x + y^2 = 0$$, so $$x = -y^2$$:

$$4y^4 + 4y^2 - 4y - 3 = 0$$

We need $$y^4 + y^2 - y$$. From the equation:

$$4(y^4 + y^2 - y) = 3$$

$$y^4 + y^2 - y = \frac{3}{4}$$

The correct answer is Option 3: $$\frac{3}{4}$$.

Given: $$\left|\frac{z - 2i}{z + i}\right| = 2$$

Let $$z = x + iy$$.

$$|z - 2i|^2 = 4|z + i|^2$$

$$x^2 + (y-2)^2 = 4[x^2 + (y+1)^2]$$

$$x^2 + y^2 - 4y + 4 = 4x^2 + 4y^2 + 8y + 4$$

$$0 = 3x^2 + 3y^2 + 12y$$

$$x^2 + y^2 + 4y = 0$$

$$x^2 + (y+2)^2 = 4$$

This is a circle with centre $$(0, -2)$$ and radius 2.

The correct answer is Option 4: (0, -2).

We are given $$z_1 = 2 + 3i$$ and $$z_2 = 3 + 4i$$ and need to determine the set $$S = \{z \in \mathbb{C} : |z - z_1|^2 - |z - z_2|^2 = |z_1 - z_2|^2\}$$.

To begin, compute $$|z_1 - z_2|^2$$. We have $$z_1 - z_2 = (2 + 3i) - (3 + 4i) = -1 - i$$ and therefore $$|z_1 - z_2|^2 = (-1)^2 + (-1)^2 = 1 + 1 = 2$$.

Next, let $$z = x + iy$$ and expand the squared distances. Using $$|z - a|^2 = (x - \operatorname{Re}(a))^2 + (y - \operatorname{Im}(a))^2$$, it follows that $$|z - z_1|^2 = (x - 2)^2 + (y - 3)^2 = x^2 - 4x + 4 + y^2 - 6y + 9$$ and $$|z - z_2|^2 = (x - 3)^2 + (y - 4)^2 = x^2 - 6x + 9 + y^2 - 8y + 16$$.

Subtracting these expressions gives
$$(x^2 - 4x + 4 + y^2 - 6y + 9) - (x^2 - 6x + 9 + y^2 - 8y + 16)$$
$$= -4x + 6x + 4 - 9 - 6y + 8y + 9 - 16$$
$$= 2x + 2y - 12$$.

Setting this equal to $$|z_1 - z_2|^2 = 2$$ leads to $$2x + 2y - 12 = 2$$, so $$2x + 2y = 14$$ and $$x + y = 7$$. This is the equation of a straight line.

The x-intercept occurs when $$y = 0$$, giving $$x = 7$$, and the y-intercept occurs when $$x = 0$$, giving $$y = 7$$. The sum of these intercepts is $$7 + 7 = 14$$.

Therefore the correct answer is Option A.

Let $$a = p + qi$$ where $$p = \text{Re}(a)$$ and $$q = \text{Im}(a)$$. Let $$z = x + yi$$.

Simplify $$\text{Re}(a + \bar{z})$$.

$$a + \bar{z} = (p + qi) + (x - yi) = (p + x) + (q - y)i$$

$$\text{Re}(a + \bar{z}) = p + x$$

Simplify $$\text{Im}(\bar{a} + z)$$.

$$\bar{a} + z = (p - qi) + (x + yi) = (p + x) + (y - q)i$$

$$\text{Im}(\bar{a} + z) = y - q$$

Write the conditions for sets $$A$$ and $$B$$.

Set $$A$$: $$\text{Re}(a + \bar{z}) > \text{Im}(\bar{a} + z)$$, which gives $$p + x > y - q$$, i.e., $$x - y + (p + q) > 0$$.

Set $$B$$: $$\text{Re}(a + \bar{z}) < \text{Im}(\bar{a} + z)$$, which gives $$p + x < y - q$$, i.e., $$x - y + (p + q) < 0$$.

Check statement (S1).

Given $$\text{Re}(a) > 0$$ and $$\text{Im}(a) > 0$$, so $$p > 0$$ and $$q > 0$$.

For a real number $$z = x$$ (where $$y = 0$$), the condition for $$A$$ becomes: $$x + (p + q) > 0$$, i.e., $$x > -(p + q)$$.

Since $$p + q > 0$$, we have $$-(p + q) < 0$$. So for any real $$x \leq -(p + q)$$, the condition fails.

Therefore, $$A$$ does NOT contain all real numbers. (S1) is FALSE.

Check statement (S2).

Given $$\text{Re}(a) < 0$$ and $$\text{Im}(a) < 0$$, so $$p < 0$$ and $$q < 0$$.

For a real number $$z = x$$ (where $$y = 0$$), the condition for $$B$$ becomes: $$x + (p + q) < 0$$, i.e., $$x < -(p + q) = |p| + |q| > 0$$.

For any real $$x \geq |p| + |q|$$, the condition fails.

Therefore, $$B$$ does NOT contain all real numbers. (S2) is FALSE.

Since both (S1) and (S2) are false, the correct answer is Option D: Both are false.

To find the correct statement regarding the relationship between the curves $$C_1$$ and $$C_2$$, we derive the Cartesian equation of the locus curve $$C_2$$ from the given circle $$C_1$$ and perform a direct geometrical comparison.

Let $$z \in \mathbf{C}$$ be a point on the curve $$C_1$$ defined by:

$$|z| = 4$$

In polar form, any point on this circle can be represented as:

$$z = 4e^{i\theta} = 4(\cos\theta + i\sin\theta)$$

---

Step 1: Derive the equation of the curve $$C_2$$

Let the points on the locus curve $$C_2$$ be represented by the complex variable $$w = x + iy$$. Given the transformation:

$$w = z + \frac{1}{z}$$

Substituting the polar form of $$z$$:

$$x + iy = 4e^{i\theta} + \frac{1}{4e^{i\theta}} = 4e^{i\theta} + \frac{1}{4}e^{-i\theta}$$

Using Euler's formula ($$e^{\pm i\theta} = \cos\theta \pm i\sin\theta$$), we expand the terms:

$$x + iy = 4(\cos\theta + i\sin\theta) + \frac{1}{4}(\cos\theta - i\sin\theta)$$

Separating this into real and imaginary parts gives:

$$x = \left(4 + \frac{1}{4}\right)\cos\theta = \frac{17}{4}\cos\theta \implies \cos\theta = \frac{4x}{17}$$

$$y = \left(4 - \frac{1}{4}\right)\sin\theta = \frac{15}{4}\sin\theta \implies \sin\theta = \frac{4y}{15}$$

Eliminating the parameter $$\theta$$ using the identity $$\cos^2\theta + \sin^2\theta = 1$$:

$$\left(\frac{4x}{17}\right)^2 + \left(\frac{4y}{15}\right)^2 = 1 \implies \frac{x^2}{\left(\frac{17}{4}\right)^2} + \frac{y^2}{\left(\frac{15}{4}\right)^2} = 1$$

Thus, the curve $$C_2$$ is an ellipse with a semi-major axis $$a = \frac{17}{4}$$ and a semi-minor axis $$b = \frac{15}{4}$$.

---

Step 2: Geometrical Comparison of Dimensions

The curve $$C_1$$ is a circle centered at the origin with radius $$R = 4$$.

The curve $$C_2$$ is an ellipse centered at the origin with semi-major axis $$a = \frac{17}{4} = 4.25$$ and semi-minor axis $$b = \frac{15}{4} = 3.75$$.

Comparing these dimensions directly:

$$b < R < a$$

$$3.75 < 4 < 4.25$$

---

Step 3: Conclusion

Since both curves are symmetric and centered at the origin, the fact that the circle's radius $$R$$ lies strictly between the semi-minor axis $$b$$ and the semi-major axis $$a$$ of the ellipse ensures that the two curves must cross each other.

Specifically, the circle extends further out than the ellipse along the y-axis ($$R > b$$), while the ellipse extends further out than the circle along the x-axis ($$a > R$$). This dimensional layout forces the curves to cross each other exactly once in each of the four quadrants.

Therefore, the correct statement is that the curves $$C_1$$ and $$C_2$$ intersect at 4 points.

For two non-zero complex numbers $$z_1$$ and $$z_2$$, let them be defined as:

$$z_1 = x_1 + i y_1$$

$$z_2 = x_2 + i y_2$$

Since $$z_1$$ and $$z_2$$ are non-zero, their real and imaginary parts cannot be zero at the same time.

Step 1: Apply the first condition

$$\text{Re}(z_1 + z_2) = 0$$

$$x_1 + x_2 = 0$$

$$x_2 = -x_1$$

Step 2: Apply the second condition

$$\text{Re}(z_1 z_2) = 0$$

$$\text{Re}((x_1 + i y_1)(x_2 + i y_2)) = 0$$

$$x_1 x_2 - y_1 y_2 = 0$$

Step 3: Substitute $$x_2 = -x_1$$ into the product equation

$$x_1(-x_1) - y_1 y_2 = 0$$

$$-x_1^2 - y_1 y_2 = 0$$

$$y_1 y_2 = -x_1^2$$

Step 4: Analyze the signs

If $$x_1 = 0$$, then $$x_2 = 0$$. This forces $$y_1 y_2 = 0$$.

If $$y_1 = 0$$, then $$z_1 = 0$$, which violates the non-zero condition.

Therefore, $$x_1 \neq 0$$, which means $$x_1^2 > 0$$.

Consequently, $$y_1 y_2 = -x_1^2 < 0$$.

Since the product of their imaginary parts is strictly negative ($$y_1 y_2 < 0$$), the imaginary parts must have opposite signs.

Possible Cases:

Case 1: $$\text{Im}(z_1) < 0$$ and $$\text{Im}(z_2) > 0$$ (Statement B)

Case 2: $$\text{Im}(z_1) > 0$$ and $$\text{Im}(z_2) < 0$$ (Statement C)

Correct answer is (B) and (C).

Consider $$z = \alpha + i\beta$$ and impose the condition $$|z + 2| = z + 4(1 + i)$$. Since the left side is real, the imaginary part of the right side must be zero. Writing the imaginary part gives $$\text{Im}(z + 4 + 4i) = \beta + 4 = 0$$, which leads to $$\beta = -4$$.

Next, equating the real parts yields $$|z + 2| = \alpha + 4$$, so $$\sqrt{(\alpha + 2)^2 + \beta^2} = \alpha + 4$$.

Squaring both sides and using the fact that $$\alpha + 4 \ge 0$$ gives $$(\alpha + 2)^2 + 16 = (\alpha + 4)^2$$. Expanding produces $$\alpha^2 + 4\alpha + 4 + 16 = \alpha^2 + 8\alpha + 16$$, which simplifies to $$4\alpha + 20 = 8\alpha + 16$$. Solving this equation leads to $$\alpha = 1$$.

With $$\alpha = 1$$ and $$\beta = -4$$ we find $$\alpha + \beta = -3$$ and $$\alpha\beta = -4$$. A quadratic equation having roots $$-3$$ and $$-4$$ can be written as $$x^2 - (-3 - 4)x + (-3)(-4) = 0$$, which simplifies to $$x^2 + 7x + 12 = 0$$.

The required equation is $$x^2 + 7x + 12 = 0$$.

Given: $$\text{Re}\left(\frac{z-\bar{z}+z\bar{z}}{2-3z+5\bar{z}}\right)$$ where $$\text{Re}(z) = 3$$.

Let $$z = 3 + iy$$. Then $$\bar{z} = 3 - iy$$.

Numerator: $$z - \bar{z} + z\bar{z} = 2iy + (9+y^2) = (9+y^2) + 2iy$$.

Denominator: $$2 - 3(3+iy) + 5(3-iy) = 2 - 9 - 3iy + 15 - 5iy = 8 - 8iy$$.

$$\frac{(9+y^2)+2iy}{8-8iy} = \frac{[(9+y^2)+2iy](8+8iy)}{64(1+y^2)}$$

Real part of the numerator product: $$8(9+y^2) + 16i^2y^2 = 8(9+y^2) - 16y^2 = 72 - 8y^2$$.

$$\text{Re} = \frac{72-8y^2}{64(1+y^2)} = \frac{9-y^2}{8(1+y^2)}$$

Let $$f(y) = \frac{9-y^2}{8(1+y^2)}$$. Setting $$f'(y) = 0$$ gives $$y = 0$$ (a maximum). At $$y = 0$$: $$f = 9/8$$. As $$|y| \to \infty$$: $$f \to -1/8$$.

Range of $$f$$: $$(-1/8, 9/8]$$. So $$\alpha = -1/8$$, $$\beta = 9/8$$.

$$24(\beta - \alpha) = 24(9/8 + 1/8) = 24 \times 10/8 = 30$$.

The correct answer is Option 3: $$30$$.

We need to find a possible value of $$\frac{1+[a]}{4b}$$ given the conditions on $$a$$ and $$b$$.

1. $$ab < 0$$ ($$a$$ and $$b$$ have opposite signs)

2. $$\left|\frac{1+ai}{b+i}\right| = 1$$ (unit modulus)

3. $$a + ib$$ lies on the circle $$|z - 1| = |2z|$$

Use the unit modulus condition.

$$\left|\frac{1+ai}{b+i}\right| = 1 \implies |1+ai| = |b+i|$$

$$\sqrt{1 + a^2} = \sqrt{b^2 + 1}$$

$$1 + a^2 = b^2 + 1$$

$$a^2 = b^2$$

$$|a| = |b|$$

Since $$ab < 0$$, we have $$a = -b$$ (they have opposite signs).

Use the circle condition.

The point $$z = a + ib$$ lies on $$|z - 1| = |2z|$$:

$$|a + ib - 1| = |2(a + ib)|$$

$$|(a-1) + ib| = |2a + 2ib|$$

$$(a-1)^2 + b^2 = 4a^2 + 4b^2$$

Since $$a = -b$$, substitute $$b = -a$$:

$$(a-1)^2 + a^2 = 4a^2 + 4a^2$$

$$a^2 - 2a + 1 + a^2 = 8a^2$$

$$2a^2 - 2a + 1 = 8a^2$$

$$6a^2 + 2a - 1 = 0$$

Solve the quadratic.

$$a = \frac{-2 \pm \sqrt{4 + 24}}{12} = \frac{-2 \pm \sqrt{28}}{12} = \frac{-2 \pm 2\sqrt{7}}{12} = \frac{-1 \pm \sqrt{7}}{6}$$

So $$a = \frac{-1 + \sqrt{7}}{6} \approx \frac{-1 + 2.646}{6} \approx \frac{1.646}{6} \approx 0.274$$

or $$a = \frac{-1 - \sqrt{7}}{6} \approx \frac{-3.646}{6} \approx -0.608$$

Since $$ab < 0$$ and $$b = -a$$:

If $$a > 0$$, then $$b < 0$$, and $$ab = -a^2 < 0$$ (valid)

If $$a < 0$$, then $$b > 0$$, and $$ab = -a^2 < 0$$ (valid)

Both values are valid.

Calculate $$\frac{1+[a]}{4b}$$.

Case 1: $$a \approx 0.274$$, so $$[a] = 0$$ and $$b = -a \approx -0.274$$

$$\frac{1 + 0}{4 \times (-0.274)} = \frac{1}{-1.096} \approx -0.912$$

Case 2: $$a \approx -0.608$$, so $$[a] = -1$$ and $$b = -a \approx 0.608$$

$$\frac{1 + (-1)}{4 \times 0.608} = \frac{0}{2.432} = 0$$

In Case 2, the value is exactly $$0$$.

The correct answer is Option A: $$0$$.

We need to find the set $$A = \left\{\theta \in (0, 2\pi) : \frac{1 + 2i\sin\theta}{1 - i\sin\theta} \text{ is purely imaginary}\right\}$$ and then compute the sum of its elements.

To simplify, rationalize the expression.

Multiply the numerator and denominator by the conjugate of the denominator:

$$\frac{1 + 2i\sin\theta}{1 - i\sin\theta} \times \frac{1 + i\sin\theta}{1 + i\sin\theta} = \frac{(1 + 2i\sin\theta)(1 + i\sin\theta)}{(1)^2 + (\sin\theta)^2}$$

Expanding the numerator.

$$= \frac{1 + i\sin\theta + 2i\sin\theta + 2i^2\sin^2\theta}{1 + \sin^2\theta}$$

$$= \frac{(1 - 2\sin^2\theta) + i(3\sin\theta)}{1 + \sin^2\theta}$$

Now apply the condition for purely imaginary.

For the expression to be purely imaginary, the real part must be zero and the imaginary part must be non-zero:

Real part = 0:

$$\frac{1 - 2\sin^2\theta}{1 + \sin^2\theta} = 0$$

$$1 - 2\sin^2\theta = 0$$

$$\sin^2\theta = \frac{1}{2}$$

$$\sin\theta = \pm\frac{1}{\sqrt{2}}$$

Imaginary part ≠ 0:

$$\frac{3\sin\theta}{1 + \sin^2\theta} \neq 0$$

This requires $$\sin\theta \neq 0$$, which is satisfied for all solutions above.

Now, find all values of $$\theta$$ in $$(0, 2\pi)$$.

From $$\sin\theta = \frac{1}{\sqrt{2}}$$: $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}$$

From $$\sin\theta = -\frac{1}{\sqrt{2}}$$: $$\theta = \frac{5\pi}{4}, \frac{7\pi}{4}$$

So $$A = \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\right\}$$

Now, compute the sum.

$$\frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi$$

The sum of the elements in $$A$$ is $$4\pi$$, which corresponds to Option A.

Now, find $$w_1$$.

Rotation of $$z_1 = 5 + 4i$$ through 90° anticlockwise is achieved by multiplying by $$i$$:

$$w_1 = i \cdot z_1 = i(5 + 4i) = 5i + 4i^2 = -4 + 5i$$

Now, find $$w_2$$.

Rotation of $$z_2 = 3 + 5i$$ through 90° clockwise is achieved by multiplying by $$-i$$:

$$w_2 = -i \cdot z_2 = -i(3 + 5i) = -3i - 5i^2 = 5 - 3i$$

Now, find $$w_1 - w_2$$.

$$w_1 - w_2 = (-4 + 5i) - (5 - 3i) = -9 + 8i$$

Now, find the principal argument.

The point $$-9 + 8i$$ lies in the second quadrant (negative real, positive imaginary).

$$\arg(w_1 - w_2) = \pi - \tan^{-1}\left(\frac{8}{9}\right)$$

Let $$\theta = \frac{2\pi}{9}$$. Note that $$\sin\theta = \cos(\pi/2 - \theta)$$ and $$\cos\theta = \sin(\pi/2 - \theta)$$.

$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}$$

Let $$a = 1 + \sin\theta$$, $$b = \cos\theta$$. Then:

$$\frac{a + ib}{a - ib} = \frac{(a+ib)^2}{a^2+b^2}$$

Note: $$\pi/2 - \theta = \pi/2 - 2\pi/9 = 5\pi/18$$

$$1 + \sin\theta = 1 + \cos(5\pi/18)$$, $$\cos\theta = \sin(5\pi/18)$$

Using $$1 + \cos\phi = 2\cos^2(\phi/2)$$:

$$a = 2\cos^2(5\pi/36)$$, $$b = 2\sin(5\pi/36)\cos(5\pi/36)$$

$$\frac{a + ib}{a - ib}$$ = write in polar form. $$a + ib = 2\cos(5\pi/36)[\cos(5\pi/36) + i\sin(5\pi/36)]$$

$$= 2\cos(5\pi/36) \cdot e^{i \cdot 5\pi/36}$$

Similarly $$a - ib = 2\cos(5\pi/36) \cdot e^{-i \cdot 5\pi/36}$$

$$\frac{a+ib}{a-ib} = e^{i \cdot 10\pi/36} = e^{i \cdot 5\pi/18}$$

Cubing: $$\left(\frac{a+ib}{a-ib}\right)^3 = e^{i \cdot 15\pi/18} = e^{i \cdot 5\pi/6}$$

$$= \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{i}{2} = \frac{1}{2}(-\sqrt{3} + i) = \frac{-1}{2}(\sqrt{3} - i)$$

The correct answer is Option 3: $$\frac{-1}{2}(\sqrt{3} - i)$$.

Given $$S = \left\{z \in \mathbb{C} - \{i, 2i\} : \dfrac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R}\right\}$$ and $$\alpha - \dfrac{13}{11}i \in S$$ with $$\alpha \in \mathbb{R} - \{0\}$$.

Factor the expressions.

$$z^2 + 8iz - 15 = (z + 3i)(z + 5i)$$

$$z^2 - 3iz - 2 = (z - i)(z - 2i)$$

Substitute $$z = \alpha - \frac{13}{11}i$$.

Computing the numerator $$N = z^2 + 8iz - 15$$:

$$z^2 = \alpha^2 - \frac{26\alpha i}{11} - \frac{169}{121}$$

$$8iz = 8\alpha i + \frac{104}{11}$$

$$N = \left(\alpha^2 - \frac{169}{121} + \frac{104}{11} - 15\right) + i\left(-\frac{26\alpha}{11} + 8\alpha\right)$$

$$= \left(\alpha^2 - \frac{840}{121}\right) + i \cdot \frac{62\alpha}{11}$$

Computing the denominator $$D = z^2 - 3iz - 2$$:

$$-3iz = -3\alpha i - \frac{39}{11}$$

$$D = \left(\alpha^2 - \frac{169}{121} - \frac{39}{11} - 2\right) + i\left(-\frac{26\alpha}{11} - 3\alpha\right)$$

$$= \left(\alpha^2 - \frac{840}{121}\right) + i \cdot \left(-\frac{59\alpha}{11}\right)$$

Apply the condition for $$N/D$$ to be real.

$$\frac{N}{D} \in \mathbb{R} \iff \text{Im}(N \cdot \overline{D}) = 0$$

Let $$R = \alpha^2 - \frac{840}{121}$$. Then $$N = R + i\frac{62\alpha}{11}$$ and $$\overline{D} = R + i\frac{59\alpha}{11}$$.

$$\text{Im}(N \cdot \overline{D}) = R \cdot \frac{59\alpha}{11} + \frac{62\alpha}{11} \cdot R = R \cdot \frac{121\alpha}{11} = 11R\alpha$$

Setting this to zero: since $$\alpha \neq 0$$, we need $$R = 0$$:

$$\alpha^2 = \frac{840}{121}$$

Compute the answer.

$$242\alpha^2 = 242 \times \frac{840}{121} = 2 \times 840 = 1680$$

The answer is $$1680$$.

Given $$\alpha, \beta, z \in \mathbb{C}$$, $$\lambda > 1$$, and $$\sqrt{\lambda - 1}$$ is the radius of the circle $$|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$$.

Expanding the left side:

$$|z|^2 - 2\text{Re}(z\bar{\alpha}) + |\alpha|^2 + |z|^2 - 2\text{Re}(z\bar{\beta}) + |\beta|^2 = 2\lambda$$

$$2|z|^2 - 2\text{Re}\left(z(\bar{\alpha} + \bar{\beta})\right) + |\alpha|^2 + |\beta|^2 = 2\lambda$$

Completing the square by writing $$z = w + \frac{\alpha + \beta}{2}$$:

$$\left|z - \frac{\alpha + \beta}{2}\right|^2 = \lambda - \frac{|\alpha - \beta|^2}{4}$$

This is a circle with center $$\frac{\alpha + \beta}{2}$$ and radius:

$$r = \sqrt{\lambda - \frac{|\alpha - \beta|^2}{4}}$$

Given that the radius equals $$\sqrt{\lambda - 1}$$:

$$\sqrt{\lambda - \frac{|\alpha - \beta|^2}{4}} = \sqrt{\lambda - 1}$$

Squaring both sides:

$$\lambda - \frac{|\alpha - \beta|^2}{4} = \lambda - 1$$

$$\frac{|\alpha - \beta|^2}{4} = 1$$

$$|\alpha - \beta| = 2$$

The answer is $$2$$.

$$z(\alpha - z) - \bar{z}(\bar{\alpha} - \bar{z}) = -112i$$

$$2i\operatorname{Im}\big(z(\alpha - z)\big) = -112i \implies \operatorname{Im}\big(z(\alpha - z)\big) = -56$$

$$z(\alpha - z) = (x + iy)\big((8 - x) - i(14 + y)\big)$$

$$y(8 - x) - x(14 + y) = -56$$

$$8y - xy - 14x - xy = -56 \implies xy + 7x - 4y - 28 = 0$$

$$(x - 4)(y + 7) = 0 \implies x = 4 \text{ or } y = -7 \quad \text{--- (1)}$$

$$x^2 + (y + 3)^2 = 16 \quad \text{--- (2)}$$

Substituting $$x = 4$$ and $$y = 7$$: 

$$\implies \mathbf{(4, -3)}$$

$$ \implies \mathbf{(0, -7)}$$

$$\sum \big(\operatorname{Re}(z) - \operatorname{Im}(z)\big) = [4 - (-3)] + [0 - (-7)] = 7 + 7 = 14$$

We are given $$z = 1 + i$$ and asked to find $$\frac{12}{\pi}\arg(z_1)$$ where $$z_1 = \frac{1 + i\bar{z}}{\bar{z}(1-z) + \frac{1}{z}}$$. To begin, we note that $$\bar{z} = 1 - i$$, so $$1 + i\bar{z} = 1 + i(1 - i) = 1 + i - i^2 = 1 + i + 1 = 2 + i$$.

Next, we calculate the denominator. First, $$\bar{z}(1 - z) = (1 - i)(1 - 1 - i) = (1 - i)(-i) = -i + i^2 = -1 - i$$, and $$\frac{1}{z} = \frac{1}{1 + i} = \frac{1 - i}{2}$$. Adding these gives $$-1 - i + \frac{1 - i}{2} = \frac{-2 - 2i + 1 - i}{2} = \frac{-1 - 3i}{2}$$.

Therefore, $$z_1 = \frac{2 + i}{\frac{-1 - 3i}{2}} = \frac{2(2 + i)}{-1 - 3i} = \frac{4 + 2i}{-1 - 3i}$$. Multiplying numerator and denominator by the conjugate of the denominator yields $$z_1 = \frac{(4 + 2i)(-1 + 3i)}{(-1 - 3i)(-1 + 3i)} = \frac{-4 + 12i - 2i + 6i^2}{1 + 9} = \frac{-4 + 10i - 6}{10} = \frac{-10 + 10i}{10} = -1 + i\,. $$

Since $$\arg(-1 + i) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$, it follows that $$\frac{12}{\pi} \times \frac{3\pi}{4} = \frac{12 \times 3}{4} = 9$$. Thus, the answer is $$\boxed{9}$$.

Let $$z$$ be written in polar form as $$z = r\,e^{i\theta} \;(r \gt 0,\, -\pi \lt \theta \le \pi)$$. Then $$\bar z = r\,e^{-i\theta}$$, so

$$ (\bar z)^2 + \frac{1}{z^{2}} = r^{2}e^{-i2\theta} + \frac{1}{r^{2}}e^{-i2\theta} = \bigl(r^{2}+r^{-2}\bigr)\,e^{-i2\theta}. $$

Put $$s = r^{2}+r^{-2} \; (>2).$$ Writing $$e^{-i2\theta} = \cos 2\theta - i\sin 2\theta,$$ we get

$$ (\bar z)^2 + \frac{1}{z^{2}} = s\cos2\theta \;-\; i\,s\sin2\theta. $$

The real part $$s\cos2\theta$$ and the imaginary part $$-\,s\sin2\theta$$ are both required to be integers. Hence there exist integers $$A,B$$ such that

$$ s\cos2\theta = A,\qquad -s\sin2\theta = B. $$

Using $$\cos^{2}2\theta+\sin^{2}2\theta = 1$$, we obtain

$$ \frac{A^{2}+B^{2}}{s^{2}} = 1 \;\;\Longrightarrow\;\; s^{2}=A^{2}+B^{2}. \tag{-1} $$

Thus $$s=\sqrt{A^{2}+B^{2}}$$ must itself be the square root of a sum of two perfect squares.

Because $$s = r^{2}+r^{-2},$$ we have the quadratic in $$r^{2}$$:

$$ (r^{2})^{2} - s\,r^{2} + 1 = 0 \;\;\Longrightarrow\;\; r^{2} = \frac{s \pm \sqrt{s^{2}-4}}{2}. $$

Taking the positive root and then $$r=\sqrt{r^{2}}$$ gives the admissible moduli $$|z|$$ for every allowed $$s$$.

We now test the four given options.

Let $$r = \Bigl(\dfrac{43+3\sqrt{205}}{2}\Bigr)^{1/4}.$$ Then

$$ r^{2} = \sqrt{\dfrac{43+3\sqrt{205}}{2}} = y \quad(\text{say}). $$

Compute $$s = y+\dfrac{1}{y}$$:

$$\begin{aligned} y^{2} &= \dfrac{43+3\sqrt{205}}{2},\\[4pt] \frac{1}{y^{2}} &= \dfrac{2}{43+3\sqrt{205}},\\[4pt] y^{2}+\frac{1}{y^{2}} &= \dfrac{43+3\sqrt{205}}{2} + \dfrac{2}{43+3\sqrt{205}} = 43, \\[4pt] s &= y+\frac{1}{y} = \sqrt{\,y^{2}+\frac{1}{y^{2}}+2\,} = \sqrt{43+2} = \sqrt{45} = 3\sqrt5. \end{aligned}$$

Because $$s^{2} = 45 = 6^{2}+3^{2},$$ equation $$-(1)$$ is satisfied with $$A=6,\;B=3.$$ Both parts of $$(\bar z)^2 + 1/z^{2}$$ can therefore be integral. Hence Option A gives a valid modulus.

Take $$r = \Bigl(\dfrac{7+\sqrt{33}}{4}\Bigr)^{1/4}.$$ Here

$$ y^{2} = \dfrac{7+\sqrt{33}}{4},\qquad y^{2}+\dfrac{1}{y^{2}} = \dfrac72,\qquad s = \sqrt{\dfrac72 + 2} = \sqrt{\dfrac{11}{2}}. $$

Then $$s^{2} = \dfrac{11}{2}$$ is not an integer, so $$A,B$$ cannot be integers. Option B is not possible.

Take $$r = \Bigl(\dfrac{9+\sqrt{65}}{4}\Bigr)^{1/4}.$$ Similarly we get

$$ y^{2} = \dfrac{9+\sqrt{65}}{4},\qquad y^{2}+\dfrac{1}{y^{2}} = \dfrac92,\qquad s^{2} = \dfrac{13}{2}, $$

which is not an integer. Option C is not possible.

Take $$r = \Bigl(\dfrac{7+\sqrt{13}}{6}\Bigr)^{1/4}.$$ We find

$$ y^{2} = \dfrac{7+\sqrt{13}}{6},\qquad y^{2}+\dfrac{1}{y^{2}} = \dfrac73,\qquad s^{2} = \dfrac{13}{3}, $$

again non-integral, so integers $$A,B$$ do not exist. Option D is not possible.

Therefore the only modulus that fulfils the given condition is

$$|z| = \Bigl(\dfrac{43 + 3\sqrt{205}}{2}\Bigr)^{1/4}.$$

Option A which is: $$\left(\dfrac{43 + 3\sqrt{205}}{2}\right)^{1/4}$$

Let the given quotient be a real number and denote it by a real constant $$k$$:

$$\frac{\,2 + 3z + 4z^{2}\,}{\,2 - 3z + 4z^{2}\,}=k,\qquad k\in\mathbb{R}$$

Cross-multiplying gives a quadratic equation in $$z$$:

$$2 + 3z + 4z^{2}=k\,(2 - 3z + 4z^{2})$$

Bring all terms to the left:

$$4z^{2}(1-k)+3z(1+k)+2(1-k)=0 \qquad -(1)$$

The coefficients $$4(1-k),\;3(1+k),\;2(1-k)$$ are real because $$k$$ is real. Equation $$(1)$$ is therefore a quadratic with real coefficients.

Since $$z$$ has a non-zero imaginary part, its complex conjugate $$\bar z$$ is also a root of the same quadratic. Thus the two roots of $$(1)$$ are $$z$$ and $$\bar z$$.

For a quadratic $$az^{2}+bz+c=0$$, the product of the roots equals $$\dfrac{c}{a}$$ (Vieta’s formula). Comparing $$(1)$$ with $$az^{2}+bz+c=0$$ we have

$$a=4(1-k),\quad b=3(1+k),\quad c=2(1-k)$$

Hence

$$z\bar z=\frac{c}{a}=\frac{2(1-k)}{4(1-k)}=\frac12$$

But $$z\bar z=|z|^{2}$$, so

$$|z|^{2}=\frac12$$

Therefore the required value is 0.50.

Let $$z = x + iy$$, where $$x, y \in \mathbb{R}$$. Then $$\bar{z} = x - iy$$ and $$z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$$.

Substitute these in the given equation
$$\bar{z} - z^2 = i\bigl(\bar{z} + z^2\bigr)$$

Left side (LHS):
$$(x - iy) - (x^2 - y^2 + 2ixy) = (x - x^2 + y^2) + i(-y - 2xy)$$

Right side (RHS):
$$i\!\Bigl[(x - iy) + (x^2 - y^2 + 2ixy)\Bigr]$$ $$= i\bigl(x + x^2 - y^2\bigr) + i(i)\bigl(-y + 2xy\bigr)$$ $$= i\bigl(x + x^2 - y^2\bigr) - \bigl(-y + 2xy\bigr)$$ $$= (\,y - 2xy\,) + i\bigl(x + x^2 - y^2\bigr)$$

Equate real and imaginary parts:

Real part:
$$x - x^2 + y^2 = y - 2xy \quad -(1)$$

Imaginary part:
$$-y - 2xy = x + x^2 - y^2 \quad -(2)$$

Rewrite both relations in a common format:

$$(1)\;:\; -x^2 + x + 2xy + y^2 - y = 0$$ $$(2)\;:\; -x^2 - 2xy - x + y^2 - y = 0$$

Subtract $$(2)$$ from $$(1)$$ to eliminate the quadratic terms:

$$(1) - (2):\; 2x + 4xy = 0 \;\Longrightarrow\; 2x(1 + 2y) = 0$$ $$\therefore \; x = 0 \quad \text{or} \quad 1 + 2y = 0$$

Put $$x = 0$$ in $$(1)$$:

$$0^2 - 0 + y^2 - y = 0 \;\Longrightarrow\; y(y - 1) = 0$$

Hence $$y = 0 \quad\text{or}\quad y = 1$$.

Solutions:

$$z = 0 \quad\text{and}\quad z = i$$

Put $$y = -\frac12$$ in $$(1)$$:

$$-x^2 + x + 2x\!\left(-\frac12\right) + \left(-\frac12\right)^2 - \left(-\frac12\right) = 0$$ $$-x^2 + x - x + \frac14 + \frac12 = 0$$ $$-x^2 + \frac34 = 0 \;\Longrightarrow\; x^2 = \frac34$$ $$\therefore\; x = \pm\frac{\sqrt3}{2}$$

Solutions:

$$z = \frac{\sqrt3}{2} - \frac{i}{2}, \quad z = -\frac{\sqrt3}{2} - \frac{i}{2}$$

Combining both cases, the four distinct roots are
$$z = 0,\quad z = i,\quad z = \frac{\sqrt3}{2} - \frac{i}{2},\quad z = -\frac{\sqrt3}{2} - \frac{i}{2}.$$

Hence, the equation possesses 4 distinct roots.

Final Answer: 4

We need the value of $$p$$ such that the minimum of $$|z - 3\sqrt{2}| + |z - p\sqrt{2}\,i|$$ is $$5\sqrt{2}$$.

Geometric interpretation.

The expression $$|z - 3\sqrt{2}| + |z - p\sqrt{2}\,i|$$ is the sum of distances from $$z$$ to two fixed points in the complex plane:

$$A = 3\sqrt{2}$$ (on the real axis) and $$B = p\sqrt{2}\,i$$ (on the imaginary axis).

Apply the triangle inequality.

By the triangle inequality, the minimum value of $$|z - A| + |z - B|$$ equals the distance $$|A - B|$$, achieved when $$z$$ lies on the line segment $$AB$$.

Compute $$|A - B|$$.

$$|A - B| = |3\sqrt{2} - p\sqrt{2}\,i| = \sqrt{(3\sqrt{2})^2 + (p\sqrt{2})^2} = \sqrt{18 + 2p^2}$$

Solve for $$p$$.

$$\sqrt{18 + 2p^2} = 5\sqrt{2}$$

Squaring both sides:

$$18 + 2p^2 = 50 \implies 2p^2 = 32 \implies p^2 = 16 \implies p = \pm 4$$

Select from the options.

Among the given options, $$p = 4$$ is available.

The answer is Option D: $$4$$.

We need to find $$ \alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021} $$, where $$ \alpha, \beta, \gamma, \delta $$ are roots of $$ x^4 + x^3 + x^2 + x + 1 = 0 $$.

The equation $$ x^4 + x^3 + x^2 + x + 1 = 0 $$ can be written as:

$$\frac{x^5 - 1}{x - 1} = 0$$

So the roots are the primitive 5th roots of unity: $$ \omega, \omega^2, \omega^3, \omega^4 $$, where $$ \omega = e^{2\pi i/5} $$.

Each root satisfies $$ \omega^5 = 1 $$.

$$2021 = 5 \times 404 + 1$$

Therefore, $$ 2021 \mod 5 = 1 $$.

For each root $$ \omega^k $$ (where $$ k = 1, 2, 3, 4 $$):

$$(\omega^k)^{2021} = (\omega^k)^{5 \times 404 + 1} = ((\omega^k)^5)^{404} \cdot \omega^k = 1^{404} \cdot \omega^k = \omega^k$$

$$\alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021} = \omega + \omega^2 + \omega^3 + \omega^4$$

By Vieta's formulas for $$ x^4 + x^3 + x^2 + x + 1 = 0 $$, the sum of roots equals $$ -1 $$ (negative of the coefficient of $$ x^3 $$).

$$\omega + \omega^2 + \omega^3 + \omega^4 = -1$$

The answer is $$ -1 $$, which corresponds to Option D.

We have $$z = 2 + 3i$$, so $$\bar{z} = 2 - 3i$$. We need to compute $$z^5 + \bar{z}^5$$. Since $$\bar{z}^5 = \overline{z^5}$$, we know that $$z^5 + \bar{z}^5 = 2\,\text{Re}(z^5)$$. So we just need the real part of $$z^5$$.

We compute successive powers of $$z$$:

$$z^2 = (2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i$$

$$z^3 = z \cdot z^2 = (2 + 3i)(-5 + 12i) = -10 + 24i - 15i + 36i^2 = -10 + 9i - 36 = -46 + 9i$$

$$z^4 = z \cdot z^3 = (2 + 3i)(-46 + 9i) = -92 + 18i - 138i + 27i^2 = -92 - 120i - 27 = -119 - 120i$$

$$z^5 = z \cdot z^4 = (2 + 3i)(-119 - 120i) = -238 - 240i - 357i - 360i^2 = -238 - 597i + 360 = 122 - 597i$$

Now, $$\text{Re}(z^5) = 122$$, so $$z^5 + \bar{z}^5 = 2 \times 122 = 244$$.

Hence, the correct answer is Option A.

We have a complex number $$z \neq 0$$ satisfying $$\left|z - \dfrac{1}{z}\right| = 2$$, and we need to find the maximum value of $$|z|$$.

We use the reverse triangle inequality. We know that $$\left|z - \dfrac{1}{z}\right| \geq \bigg||z| - \dfrac{1}{|z|}\bigg|$$. Since $$\left|z - \dfrac{1}{z}\right| = 2$$, we get $$\bigg||z| - \dfrac{1}{|z|}\bigg| \leq 2$$.

Now let $$r = |z|$$. We need $$\left|r - \dfrac{1}{r}\right| \leq 2$$, which gives us $$-2 \leq r - \dfrac{1}{r} \leq 2$$.

We focus on the upper bound $$r - \dfrac{1}{r} \leq 2$$. Multiplying by $$r > 0$$, we get $$r^2 - 2r - 1 \leq 0$$. Using the quadratic formula, $$r \leq \dfrac{2 + \sqrt{4+4}}{2} = 1 + \sqrt{2} = \sqrt{2} + 1$$.

We need to verify this maximum is achievable. If $$z = r$$ is a positive real number with $$r = \sqrt{2}+1$$, then $$\dfrac{1}{r} = \dfrac{1}{\sqrt{2}+1} = \sqrt{2}-1$$, so $$z - \dfrac{1}{z} = (\sqrt{2}+1) - (\sqrt{2}-1) = 2$$. This satisfies the given condition, confirming the maximum is attained.

Hence, the correct answer is Option D: $$\sqrt{2}+1$$.

We need to find $$\arg z$$ for a point $$z \neq z_1$$ on the circle $$C$$ such that the line through $$z$$ and $$z_1$$ is perpendicular to the line through $$z_2$$ and $$z_3$$. The circle passes through $$z_1 = 3 + 4i$$, $$z_2 = 4 + 3i$$, and $$z_3 = 5i$$, corresponding to the points $$(3,4)$$, $$(4,3)$$, and $$(0,5)$$.

Using the general equation $$x^2 + y^2 + Dx + Ey + F = 0$$ leads to three linear equations: from $$(3,4)$$ we get $$3D + 4E + F = -25$$, from $$(4,3)$$ we have $$4D + 3E + F = -25$$, and from $$(0,5)$$ we deduce $$5E + F = -25$$. Subtracting the second equation from the first gives $$-D + E = 0$$, so $$D = E$$, and then $$F = -25 - 5E$$. Substituting back into the first equation leads to $$3E + 4E + (-25 - 5E) = -25$$, which simplifies to $$2E = 0$$, hence $$E = 0$$, followed by $$D = 0$$ and $$F = -25$$. Therefore, the circle is $$x^2 + y^2 = 25$$.

Next, the slope of the line through $$z_2$$ and $$z_3$$ is determined by the direction $$z_3 - z_2 = 5i - (4+3i) = -4 + 2i$$, which gives a slope of $$\frac{2}{-4} = -\frac{1}{2}$$. A perpendicular slope is therefore $$2$$, so the line through $$(3,4)$$ can be written as $$y - 4 = 2(x - 3)$$, or equivalently $$y = 2x - 2$$.

To find the intersection of this line with the circle, we substitute $$y = 2x - 2$$ into $$x^2 + y^2 = 25$$, which leads to
$$x^2 + (2x-2)^2 = 25$$
$$x^2 + 4x^2 - 8x + 4 = 25$$
$$5x^2 - 8x - 21 = 0$$. The quadratic formula then gives
$$x = \frac{8 \pm \sqrt{64 + 420}}{10} = \frac{8 \pm \sqrt{484}}{10} = \frac{8 \pm 22}{10}$$, hence $$x = 3$$ (which corresponds to $$z_1$$) or $$x = -\frac{7}{5}$$, and in the latter case $$y = 2\left(-\frac{7}{5}\right) - 2 = -\frac{24}{5}$$.

Therefore, the required point is $$z = -\frac{7}{5} - \frac{24}{5}i$$, which lies in the third quadrant. Since $$\tan\theta = \frac{-24/5}{-7/5} = \frac{24}{7}$$ and the point is in the third quadrant, its argument is given by
$$\arg z = \tan^{-1}\frac{24}{7} - \pi$$.

The correct answer is Option A: $$\tan^{-1}\dfrac{24}{7} - \pi$$.

$$\alpha$$ is a root of $$1 + x^2 + x^4 = 0$$. We wish to find $$\alpha^{1011} + \alpha^{2022} - \alpha^{3033}$$.

Let $$y = x^2$$ so that the equation becomes $$y^2 + y + 1 = 0$$. Its roots are $$y = \omega$$ and $$y = \omega^2$$, where $$\omega = e^{2\pi i/3}$$ is a primitive cube root of unity satisfying $$\omega^3 = 1$$.

Since $$\alpha^2 = \omega$$ (or $$\alpha^2 = \omega^2$$), it follows that $$\alpha^6 = (\alpha^2)^3 = \omega^3 = 1$$, so $$\alpha$$ is a primitive 6th root of unity.

Reducing exponents modulo 6, we write $$1011 = 168\times6 + 3$$ so $$\alpha^{1011} = \alpha^3$$, $$2022 = 337\times6 + 0$$ so $$\alpha^{2022} = \alpha^0 = 1$$, and $$3033 = 505\times6 + 3$$ so $$\alpha^{3033} = \alpha^3$$.

Therefore, $$\alpha^{1011} + \alpha^{2022} - \alpha^{3033} = \alpha^3 + 1 - \alpha^3 = 1$$. Option A: $$1$$

We are given O (origin), A = $$z_1 = 1 + 2i$$, and B = $$z_2$$ with $$\text{Re}(z_2) < 0$$, such that OAB is a right-angled isosceles triangle with OB as hypotenuse.

We start by using the Pythagorean theorem and the isosceles condition. Since OB is the hypotenuse, the right angle is at A, and OA = AB.

By the Pythagorean theorem: $$|z_2 - z_1|^2 + |z_1|^2 = |z_2|^2$$

Let $$z_2 = a + bi$$:

$$(a-1)^2 + (b-2)^2 + 5 = a^2 + b^2$$

Therefore, $$-2a - 4b + 10 = 0 \implies a + 2b = 5 \quad \cdots(1)$$

Next, applying OA = AB gives $$|z_1|^2 = |z_2 - z_1|^2$$.

Hence, $$5 = (a-1)^2 + (b-2)^2 \quad \cdots(2)$$

Then, from (1) we have $$a = 5 - 2b$$. Substituting into (2) yields

$$(4 - 2b)^2 + (b - 2)^2 = 5$$

which can be rewritten as $$4(b-2)^2 + (b-2)^2 = 5$$

Hence, $$5(b-2)^2 = 5 \implies b = 3 \text{ or } b = 1$$

If $$b = 3$$ then $$a = -1$$, which satisfies $$\text{Re}(z_2) < 0$$, whereas $$b = 1$$ leads to $$a = 3$$ and is therefore rejected.

As a result, $$z_2 = -1 + 3i$$.

Finally, verifying each option shows that for Option A, $$\arg(z_2) = \arg(-1 + 3i) = \pi - \tan^{-1}3$$, which is TRUE.

Similarly, Option B gives $$z_1 - 2z_2 = (1+2i) - 2(-1+3i) = 3 - 4i$$ so that $$\arg(3 - 4i) = -\tan^{-1}\frac{4}{3}$$, which is TRUE.

Option C yields $$|z_2| = \sqrt{1 + 9} = \sqrt{10}$$, which is TRUE.

However, Option D produces $$2z_1 - z_2 = 2(1+2i) - (-1+3i) = 3 + i$$ and hence $$|3 + i| = \sqrt{10} \neq 5$$, making it NOT TRUE.

Therefore, the correct answer is Option D.

We need to find the least value of $$|z_2 - z_1|$$ for $$z_1 \in S_1$$ and $$z_2 \in S_2$$.

Since $$S_1 = \{z_1 \in \mathbb{C} : |z_1 - 3| = \frac{1}{2}\}$$, it is a circle centred at $$(3, 0)$$ with radius $$\frac{1}{2}$$.

We have $$S_2 = \{z_2 \in \mathbb{C} : |z_2 - |z_2 + 1|| = |z_2 + |z_2 - 1||\}$$. Since $$|z_2 + 1|$$ and $$|z_2 - 1|$$ are non-negative real numbers, we set $$r_1 = |z_2 + 1|$$ and $$r_2 = |z_2 - 1|$$, and the condition becomes $$|z_2 - r_1| = |z_2 + r_2|$$.

Squaring both sides (valid since both sides are non-negative) gives $$|z_2 - r_1|^2 = |z_2 + r_2|^2$$. Letting $$z_2 = x + iy$$ leads to $$(x - r_1)^2 + y^2 = (x + r_2)^2 + y^2$$, which simplifies to $$-2xr_1 + r_1^2 = 2xr_2 + r_2^2$$. Equivalently, $$(r_1 + r_2)(r_1 - r_2) = 2x(r_1 + r_2)$$.

Since $$r_1 + r_2 = |z_2 + 1| + |z_2 - 1| > 0$$, dividing by $$(r_1 + r_2)$$ yields $$r_1 - r_2 = 2x$$, namely $$|z_2 + 1| - |z_2 - 1| = 2\text{Re}(z_2) \quad \cdots (*)$$.

Substituting $$z_2 = x + iy$$ into (*) gives $$\sqrt{(x+1)^2 + y^2} - \sqrt{(x-1)^2 + y^2} = 2x$$. Rearranging leads to $$\sqrt{(x+1)^2 + y^2} = 2x + \sqrt{(x-1)^2 + y^2}$$, requiring $$2x + \sqrt{(x-1)^2 + y^2} \geq 0$$, which holds for all $$x \geq 0$$ and also for some $$x < 0$$.

Squaring both sides again yields $$(x+1)^2 + y^2 = 4x^2 + 4x\sqrt{(x-1)^2 + y^2} + (x-1)^2 + y^2$$, which simplifies through cancellation to $$4x = 4x^2 + 4x\sqrt{(x-1)^2 + y^2}$$ and hence $$4x\left(1 - x - \sqrt{(x-1)^2 + y^2}\right) = 0$$.

One possibility is $$x = 0$$. Substituting back into (*) gives $$|iy + 1| - |iy - 1| = 0$$, and since $$\sqrt{1+y^2} = \sqrt{1+y^2}$$ holds identically, all points on the imaginary axis are in $$S_2$$.

The other possibility with $$x \neq 0$$ forces $$1 - x = \sqrt{(x-1)^2 + y^2}$$, which requires $$1 - x \geq 0$$ (so $$x \leq 1$$), and squaring gives $$(1-x)^2 = (x-1)^2 + y^2$$, hence $$y^2 = 0$$ or $$y = 0$$. For $$y = 0$$ and $$-1 \leq x \leq 1$$, one checks that $$|x+1| - |x-1| = (x+1) - (1-x) = 2x = 2x$$, so these points satisfy (*). For $$x > 1$$ or $$x < -1$$ the equality fails except at the endpoints $$x = \pm 1$$.

Therefore, $$S_2$$ consists of the entire imaginary axis together with the real line segment $$[-1, 1]$$.

Finally, since the circle $$S_1$$ is centred at $$(3, 0)$$ with radius $$\frac{1}{2}$$, the closest part of $$S_2$$ to this circle is the point $$(1, 0)$$ on the real segment. The point on $$S_1$$ nearest to $$(1, 0)$$ lies in the same direction and is $$\left(3 - \frac{1}{2}, 0\right) = \left(\frac{5}{2}, 0\right)$$, giving a minimum distance of $$\frac{5}{2} - 1 = \frac{3}{2}$$.

Therefore, the least value of $$|z_2 - z_1|$$ is $$\frac{3}{2}$$.

The correct answer is Option C: $$\frac{3}{2}$$.

We need to find the non-real roots of $$\bar{z} = iz^2$$.

Let $$z = x + iy$$, so $$\bar{z} = x - iy$$.

Then $$iz^2 = i(x + iy)^2 = i(x^2 - y^2 + 2ixy) = i(x^2 - y^2) + i^2(2xy) = -2xy + i(x^2 - y^2)$$

Equating real and imaginary parts of $$\bar{z} = iz^2$$:

$$x = -2xy \quad \cdots (1)$$

$$-y = x^2 - y^2 \quad \cdots (2)$$

From equation (1): $$x(1 + 2y) = 0$$

Case 1: $$x = 0$$

From equation (2): $$-y = 0 - y^2 \implies y^2 - y = 0 \implies y(y-1) = 0$$

So $$y = 0$$ giving $$z = 0$$ (real root, excluded), or $$y = 1$$ giving $$z = i$$ (non-real root).

Case 2: $$1 + 2y = 0 \implies y = -\frac{1}{2}$$

From equation (2): $$-\left(-\frac{1}{2}\right) = x^2 - \frac{1}{4}$$

$$\frac{1}{2} = x^2 - \frac{1}{4}$$

$$x^2 = \frac{3}{4}$$

$$x = \pm\frac{\sqrt{3}}{2}$$

This gives two non-real roots: $$z_2 = \frac{\sqrt{3}}{2} - \frac{i}{2}$$ and $$z_3 = -\frac{\sqrt{3}}{2} - \frac{i}{2}$$

The three non-real roots are:

$$z_1 = i = (0, 1)$$

$$z_2 = \left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$

$$z_3 = \left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$

These form a triangle. We use the formula:

Area $$= \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

$$= \frac{1}{2}\left|0\left(-\frac{1}{2} + \frac{1}{2}\right) + \frac{\sqrt{3}}{2}\left(-\frac{1}{2} - 1\right) + \left(-\frac{\sqrt{3}}{2}\right)\left(1 + \frac{1}{2}\right)\right|$$

$$= \frac{1}{2}\left|0 + \frac{\sqrt{3}}{2}\left(-\frac{3}{2}\right) - \frac{\sqrt{3}}{2}\left(\frac{3}{2}\right)\right|$$

$$= \frac{1}{2}\left|-\frac{3\sqrt{3}}{4} - \frac{3\sqrt{3}}{4}\right|$$

$$= \frac{1}{2} \times \frac{6\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$$

The area of the polygon is $$\dfrac{3\sqrt{3}}{4}$$.

The correct answer is Option B.

Given,

$$|z-(4+3i)|=2$$

This represents a circle with centre

$$(4,3)$$

and radius

$$2$$

Also,

$$|z|+|z-4|=6$$

represents an ellipse with foci

$$(0,0)\quad \text{and}\quad (4,0)$$

and major axis length

$$6$$

Hence,

$$2a=6\Rightarrow a=3$$

Distance between foci is

$$2c=4\Rightarrow c=2$$

Therefore,

$$b^2=a^2-c^2=9-4=5$$

So, equation of ellipse is

$$\frac{(x-2)^2}{9}+\frac{y^2}{5}=1$$

Now, the circle is

$$ (x-4)^2+(y-3)^2=4 $$

The topmost point of the ellipse is

$$\left(2,\sqrt5\right)$$

and the lowest point of the circle is

$$ (4,1) $$

Since

$$1<\sqrt5$$

the circle overlaps the ellipse.

Also, the centre of the circle lies outside the ellipse, so the circle cuts the ellipse at two distinct points.

Hence, the number of intersection points is

$$\boxed{2}$$

We are given $$ S_n = \{z \in \mathbb{C} : |z - 3 + 2i| = \frac{n}{4}\} $$ and $$ T_n = \{z \in \mathbb{C} : |z - 2 + 3i| = \frac{1}{n}\} $$.

$$ S_n $$ is a circle centered at $$ (3, -2) $$ with radius $$ r_1 = \frac{n}{4} $$, and $$ T_n $$ is a circle centered at $$ (2, -3) $$ with radius $$ r_2 = \frac{1}{n} $$.

$$d = \sqrt{(3-2)^2 + (-2+3)^2} = \sqrt{2}$$

Two circles are disjoint when $$ d > r_1 + r_2 $$, i.e., the circles are too far apart to touch each other externally. This requires:

$$\frac{n}{4} + \frac{1}{n} < \sqrt{2}$$

Multiplying through by $$ 4n $$:

$$n^2 + 4 < 4\sqrt{2} \cdot n$$

$$n^2 - 4\sqrt{2}\,n + 4 < 0$$

The roots of $$ n^2 - 4\sqrt{2}\,n + 4 = 0 $$ are:

$$n = \frac{4\sqrt{2} \pm \sqrt{32 - 16}}{2} = \frac{4\sqrt{2} \pm 4}{2} = 2\sqrt{2} \pm 2$$

So $$ n \in (2\sqrt{2} - 2, \, 2\sqrt{2} + 2) \approx (0.828, \, 4.828) $$.

The natural numbers in this interval are $$ n = 1, 2, 3, 4 $$.

$$ n = 1 $$: $$ r_1 + r_2 = 1.25 < 1.414 $$. Disjoint. ✓

$$ n = 2 $$: $$ r_1 + r_2 = 1.0 < 1.414 $$. Disjoint. ✓

$$ n = 3 $$: $$ r_1 + r_2 \approx 1.083 < 1.414 $$. Disjoint. ✓

$$ n = 4 $$: $$ r_1 + r_2 = 1.25 < 1.414 $$. Disjoint. ✓

$$ n = 5 $$: $$ r_1 + r_2 = 1.45 > 1.414 $$. Not disjoint by this condition. ✗

There are exactly 4 natural numbers for which the sum of radii is less than the distance between centers, making $$ S_n \cap T_n = \emptyset $$.

The answer is Option D.

We need to find the relationship between $$x$$ and $$y$$ when $$z = x + iy$$ satisfies $$|z| - 2 = 0$$ and $$|z - i| - |z + 5i| = 0$$.

$$|z| = 2 \Rightarrow x^2 + y^2 = 4$$

$$|z - i| = |z + 5i|$$

$$|x + i(y-1)| = |x + i(y+5)|$$

$$x^2 + (y-1)^2 = x^2 + (y+5)^2$$

$$y^2 - 2y + 1 = y^2 + 10y + 25$$

$$-2y + 1 = 10y + 25$$

$$-12y = 24$$

$$y = -2$$

$$x^2 + (-2)^2 = 4$$

$$x^2 = 0$$

$$x = 0$$

Option A: $$x + 2y - 4 = 0 + (-4) - 4 = -8 \neq 0$$

Option B: $$x^2 + y - 4 = 0 + (-2) - 4 = -6 \neq 0$$

Option C: $$x + 2y + 4 = 0 + (-4) + 4 = 0$$ $$\checkmark$$

Option D: $$x^2 - y + 3 = 0 - (-2) + 3 = 5 \neq 0$$

Therefore, the correct answer is Option C: $$x + 2y + 4 = 0$$.

We are given:

$$A = \{z \in \mathbb{C} : 1 \leqslant |z - (1+i)| \leqslant 2\}$$

$$B = \{z \in A : |z - (1-i)| = 1\}$$

Set $$A$$ is an annular region (ring) centered at $$(1, 1)$$ with inner radius 1 and outer radius 2.

Set $$B$$ consists of points that lie in $$A$$ and also on a circle centered at $$(1, -1)$$ with radius 1.

The circle $$|z - (1-i)| = 1$$ is centered at $$(1, -1)$$ with radius 1.

The distance between the centers $$(1, 1)$$ and $$(1, -1)$$ is:

$$ d = \sqrt{(1-1)^2 + (1-(-1))^2} = \sqrt{0 + 4} = 2 $$

For the circle $$|z - (1-i)| = 1$$ (center $$(1,-1)$$, radius 1), the distance from any point on this circle to $$(1,1)$$ ranges from $$d - 1 = 1$$ to $$d + 1 = 3$$.

For points in $$A$$, we need $$1 \leqslant |z - (1+i)| \leqslant 2$$.

Since points on the circle $$|z-(1-i)|=1$$ have $$|z-(1+i)|$$ ranging from 1 to 3, and $$A$$ requires $$|z-(1+i)| \in [1, 2]$$, the intersection is a continuous arc of the circle where $$1 \leqslant |z-(1+i)| \leqslant 2$$.

Since the range [1, 2] overlaps with [1, 3], there is a non-trivial arc of the circle that lies in $$A$$. This arc contains infinitely many points.

Therefore, $$B$$ is an infinite set.

The answer is Option D: is an infinite set.

We are given $$a = \alpha - i\beta$$ and the system of equations:

$$4ix + (1+i)y = 0 \quad \cdots(1)$$

$$8\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)x + \bar{a}y = 0 \quad \cdots(2)$$

Since $$\cos\frac{2\pi}{3} = -\frac{1}{2}$$ and $$\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$$, it follows that $$8\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -4 + 4\sqrt{3}\,i$$. Also, $$\bar{a} = \alpha + i\beta$$.

For the homogeneous system to have more than one solution, the determinant of the coefficient matrix must be zero:

$$\begin{vmatrix} 4i & 1+i \\ -4 + 4\sqrt{3}\,i & \alpha + i\beta \end{vmatrix} = 0$$

Expanding this determinant gives

$$4i(\alpha + i\beta) - (1+i)(-4 + 4\sqrt{3}\,i) = 0$$

In the first term, $$4i\alpha + 4i^2\beta = 4\alpha i - 4\beta$$. In the second term,

$$ (1+i)(-4 + 4\sqrt{3}\,i) = -4 + 4\sqrt{3}\,i - 4i + 4\sqrt{3}\,i^2 = -4 + 4\sqrt{3}\,i - 4i - 4\sqrt{3} = -(4 + 4\sqrt{3}) + (4\sqrt{3} - 4)i $$.

Putting these together leads to

$$(-4\beta + 4\alpha i) - (-(4 + 4\sqrt{3}) + (4\sqrt{3} - 4)i) = 0$$

which simplifies to

$$(-4\beta + 4 + 4\sqrt{3}) + (4\alpha - 4\sqrt{3} + 4)i = 0$$

Equating the real and imaginary parts to zero yields

Real part: $$-4\beta + 4 + 4\sqrt{3} = 0 \quad\Longrightarrow\quad \beta = 1 + \sqrt{3}$$

Imaginary part: $$4\alpha - 4\sqrt{3} + 4 = 0 \quad\Longrightarrow\quad \alpha = \sqrt{3} - 1$$

Finally,

$$\frac{\alpha}{\beta} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} = \frac{(\sqrt{3} - 1)}{(\sqrt{3} + 1)} \times \frac{(\sqrt{3} - 1)}{(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$

Therefore, $$\frac{\alpha}{\beta} = 2 - \sqrt{3}$$.

The correct answer is Option A: $$2 - \sqrt{3}$$.

$$\frac{1-i\sin\alpha}{1+2i\sin\alpha}$$

is purely imaginary.

Rationalizing,
$$\frac{(1-i\sin\alpha)(1-2i\sin\alpha)}{1+4\sin^2\alpha}$$

$$\frac{1-2\sin^2\alpha-3i\sin\alpha}{1+4\sin^2\alpha}$$

Purely imagina$$ry(\Rightarrow)$$
$$1-2\sin^2\alpha=0$$
$$\sin^2\alpha=\frac{1}{2}$$

Since$$(\pi<\alpha<2\pi),$$
$$\alpha=\frac{5\pi}{4},\frac{7\pi}{4}$$

Similarly, $$\frac{1+i\cos\beta}{1-2i\cos\beta}$$

purely real gives
$$\cos\beta=0$$

and in$$((\pi,2\pi)),$$
$$\beta=\frac{3\pi}{2}$$
$$Z_{\alpha\beta}=\sin2\alpha+i\cos2\beta$$

Since
$$\cos2\beta=\cos3\pi=-1$$

we get
$$Z=\sin2\alpha-i$$

Thus:

• for $$(\alpha=\frac{5\pi}{4}),(Z=1-i)$$
• for $$(\alpha=\frac{7\pi}{4}),(Z=-1-i)$$

Compute
$$iZ+\frac{1}{i\overline{Z}}$$

For (Z=1-i):
$$1+i+\frac{1}{-1+i}=\frac{1+i}{2}$$

For (Z=-1-i):
$$1-i+\frac{1}{-1-i}=\frac{1-i}{2}$$

Hence

$$\sum_{(\alpha,\beta)\in S}^{ }\left(iZ_{\alpha\beta}+\frac{1}{i\overline{Z}_{\alpha\beta}}\right)$$

$$\frac{1+i}{2}+\frac{1-i}{2}=1$$

We have $$S = \{z = x + iy : |z-1+i| \geq |z|,\; |z| < 2,\; |z+i| = |z-1|\}$$, and we need $$w = 2x + iy \in S$$ for some $$y \in \mathbb{R}$$. We must find the set of all such values of $$x$$.

Since $$w = 2x + iy \in S$$, we write $$w = u + iv$$ where $$u = 2x$$ and $$v = y$$. The conditions on $$w$$ are:

Condition 1: $$|w + i| = |w - 1|$$, i.e., $$|u + i(v+1)| = |(u-1) + iv|$$. Squaring: $$u^2 + (v+1)^2 = (u-1)^2 + v^2$$, which simplifies to $$u^2 + v^2 + 2v + 1 = u^2 - 2u + 1 + v^2$$, giving $$2v = -2u$$, so $$v = -u$$. Thus $$y = -2x$$.

Condition 2: $$|w| < 2$$, i.e., $$u^2 + v^2 < 4$$. Since $$v = -u$$, we get $$u^2 + u^2 < 4$$, so $$2u^2 < 4$$, meaning $$u^2 < 2$$, i.e., $$-\sqrt{2} < u < \sqrt{2}$$. Since $$u = 2x$$, we have $$-\dfrac{1}{\sqrt{2}} < x < \dfrac{1}{\sqrt{2}}$$.

Condition 3: $$|w - 1 + i| \geq |w|$$, i.e., $$|(u-1) + i(v+1)| \geq |u + iv|$$. Squaring: $$(u-1)^2 + (v+1)^2 \geq u^2 + v^2$$, which gives $$-2u + 1 + 2v + 1 \geq 0$$, so $$-2u + 2v + 2 \geq 0$$, i.e., $$v \geq u - 1$$. Substituting $$v = -u$$: $$-u \geq u - 1$$, so $$1 \geq 2u$$, giving $$u \leq \dfrac{1}{2}$$, i.e., $$x \leq \dfrac{1}{4}$$.

Combining: $$-\dfrac{1}{\sqrt{2}} < x \leq \dfrac{1}{4}$$.

Hence, the correct answer is Option B: $$\left(-\dfrac{1}{\sqrt{2}}, \dfrac{1}{4}\right)$$.

Given,

$$|z|=3$$

which represents the circle

$$x^2+y^2=9$$

Also,

$$\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$$

$$\Rightarrow \arg\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}$$

This represents the locus of points subtending an angle $$\frac{\pi}{4}$$ at the points

$$(-1,0)\ \text{and}\ (1,0)$$

Hence, the locus is a circle having chord length

$$PQ=2$$

Using

$$PQ=2R\sin\theta$$

we get

$$2=2R\sin\frac{\pi}{4}$$

$$R=\sqrt2$$

Let the center be $$(0,a)$$.

Then,

$$a^2+1=(\sqrt2)^2$$

$$a=1$$

So, the locus circle has center

$$(0,1)$$

and radius

$$\sqrt2$$

Now, distance between the centers of the two circles is

$$1$$

and

$$1+\sqrt2<3$$

Hence, the smaller circle lies completely inside the circle

$$|z|=3$$

Therefore, the two curves do not intersect.

Hence, no value of $$z$$ satisfies both equations.

Therefore, the correct answer is

$$\boxed{\text{Option C: Nowhere}}$$

We express $$z_1$$ and $$z_2$$ in polar form as $$z_1 = re^{i\theta_1}$$ and $$z_2 = se^{i\theta_2}$$, and since $$\bar{z}_1 = iz_2$$, it follows that $$re^{-i\theta_1} = is\cdot e^{i\theta_2} = se^{i(\theta_2 + \pi/2)}$$.

Comparing moduli gives $$r = s$$, while comparing arguments yields $$-\theta_1 = \theta_2 + \frac{\pi}{2}$$, so that $$\theta_2 = -\theta_1 - \frac{\pi}{2}$$.

Next, using the condition $$\arg\!\bigl(\tfrac{z_1}{z_2}\bigr) = \pi$$, we have $$\arg\!\bigl(\tfrac{z_1}{z_2}\bigr) = \theta_1 - \theta_2 = \theta_1 - \bigl(-\theta_1 - \tfrac{\pi}{2}\bigr) = 2\theta_1 + \tfrac{\pi}{2} = \pi,$$ which implies $$2\theta_1 = \frac{\pi}{2}$$ and hence $$\theta_1 = \frac{\pi}{4}$$.

Therefore, $$\arg(z_1) = \frac{\pi}{4}$$, corresponding to Option C.

We are given $$S = \{z \in \mathbb{C} : |z - 3| \leq 1 \text{ and } z(4 + 3i) + \bar{z}(4 - 3i) \leq 24\}$$.

Let $$z = x + iy$$. Then $$\bar{z} = x - iy$$.

The first condition $$|z - 3| \leq 1$$ describes a disk centered at $$(3, 0)$$ with radius 1.

For the second condition:

$$z(4+3i) + \bar{z}(4-3i) = (x+iy)(4+3i) + (x-iy)(4-3i)$$

$$= (4x - 3y + i(3x+4y)) + (4x - 3y - i(3x+4y))$$

$$= 2(4x - 3y) = 8x - 6y$$

So the second condition is: $$8x - 6y \leq 24$$, or $$4x - 3y \leq 12$$.

We need to find the point in $$S$$ closest to $$4i = (0, 4)$$.

The distance from $$(0, 4)$$ to the center $$(3, 0)$$ is $$\sqrt{9 + 16} = 5$$.

The closest point on the circle $$|z-3| = 1$$ to $$(0,4)$$ lies along the line from $$(3,0)$$ to $$(0,4)$$.

The unit vector from $$(3,0)$$ toward $$(0,4)$$ is $$\frac{(-3, 4)}{5}$$.

The closest point on the circle is: $$(3, 0) + 1 \cdot \frac{(-3, 4)}{5} = \left(3 - \frac{3}{5}, \frac{4}{5}\right) = \left(\frac{12}{5}, \frac{4}{5}\right)$$

Checking if this point satisfies $$4x - 3y \leq 12$$:

$$4 \cdot \frac{12}{5} - 3 \cdot \frac{4}{5} = \frac{48 - 12}{5} = \frac{36}{5} = 7.2 \leq 12$$ $$\checkmark$$

So the closest point is $$\alpha + i\beta = \frac{12}{5} + \frac{4}{5}i$$, giving $$\alpha = \frac{12}{5}$$ and $$\beta = \frac{4}{5}$$.

$$25(\alpha + \beta) = 25\left(\frac{12}{5} + \frac{4}{5}\right) = 25 \cdot \frac{16}{5} = 80$$

The correct answer is $$80$$.

We need complex numbers $$z = a + ib$$ with $$b \neq 0$$ satisfying $$z^2 = \bar{z} \cdot 2^{1-|z|}$$, and then the least $$n \in \mathbb{N}$$ such that $$z^n = (z+1)^n$$.

Writing $$z = re^{i\theta}$$ where $$r = |z|$$ and $$\theta \neq 0, \pi$$ (since $$b \neq 0$$), and $$\bar{z} = re^{-i\theta}$$, the equation becomes:

$$r^2 e^{2i\theta} = r e^{-i\theta} \cdot 2^{1-r}$$

Comparing magnitudes: $$r^2 = r \cdot 2^{1-r}$$, so $$r = 2^{1-r}$$ (dividing by $$r > 0$$). Testing $$r = 1$$: $$1 = 2^0 = 1$$. This works, and since $$f(r) = r$$ is increasing while $$g(r) = 2^{1-r}$$ is decreasing, $$r = 1$$ is the unique solution.

Comparing arguments: $$2\theta \equiv -\theta \pmod{2\pi}$$, so $$3\theta = 2k\pi$$ for some integer $$k$$, giving $$\theta = \frac{2k\pi}{3}$$.

Since $$b \neq 0$$, we need $$\theta \neq 0, \pi$$. The values $$\theta = \frac{2\pi}{3}$$ (for $$k = 1$$) and $$\theta = \frac{4\pi}{3}$$ (for $$k = 2$$) both satisfy this. These give $$z = e^{2\pi i/3}$$ and its conjugate $$z = e^{-2\pi i/3}$$.

Taking $$z = e^{2\pi i/3}$$, we compute $$z + 1$$:

$$z + 1 = 1 + \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = 1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = e^{i\pi/3}$$

Now $$z^n = (z+1)^n$$ requires $$\left(\frac{z}{z+1}\right)^n = 1$$. We compute:

$$\frac{z}{z+1} = \frac{e^{2\pi i/3}}{e^{i\pi/3}} = e^{i(2\pi/3 - \pi/3)} = e^{i\pi/3}$$

For $$e^{in\pi/3} = 1$$, we need $$\frac{n\pi}{3} = 2m\pi$$ for some positive integer $$m$$, which gives $$n = 6m$$. The least such positive integer is $$n = 6$$.

We can verify with $$z = e^{-2\pi i/3}$$: then $$z + 1 = e^{-i\pi/3}$$ and $$z/(z+1) = e^{-i\pi/3}$$, so $$e^{-in\pi/3} = 1$$ also requires $$n = 6$$.

Hence, the correct answer is 6.

Given,

$$\bar z=iz^2+z^2-z$$

Let

$$z=x+iy$$

Then,

$$\bar z=x-iy$$

Substituting,

$$x-iy=(1+i)(x+iy)^2-(x+iy)$$

Comparing real and imaginary parts,

$$2x=x^2-y^2-2xy$$

and

$$x^2-y^2+2xy=0$$

Using the second equation in the first,

$$2x=-4xy$$

$$x(1+2y)=0$$

Hence,

$$x=0\quad \text{or}\quad y=-\frac12$$

Case 1:

$$x=0$$

Then from

$$x^2-y^2+2xy=0$$

we get

$$y=0$$

Hence,

$$z=0$$

Case 2:

$$y=-\frac12$$

Substituting in

$$x^2-y^2+2xy=0$$

$$x^2-\frac14-x=0$$

$$4x^2-4x-1=0$$

$$2x-1=\pm\sqrt2$$

$$x=\frac{1\pm\sqrt2}{2}$$

Thus,

$$z=\frac{1+\sqrt2}{2}-\frac{i}{2}$$

or

$$z=\frac{1-\sqrt2}{2}-\frac{i}{2}$$

Now,

$$|0|^2=0$$

$$\left|\frac{1+\sqrt2}{2}-\frac{i}{2}\right|^2 = \frac{(1+\sqrt2)^2+1}{4}$$

$$\left|\frac{1-\sqrt2}{2}-\frac{i}{2}\right|^2 = \frac{(1-\sqrt2)^2+1}{4}$$

Therefore,

$$0+\frac{(1+\sqrt2)^2+1}{4}+\frac{(1-\sqrt2)^2+1}{4}=2$$

Hence,

$$\boxed{2}$$

Given $$z^2 + z + 1 = 0$$, find $$\left|\displaystyle\sum_{n=1}^{15}\left(z^n + (-1)^n \frac{1}{z^n}\right)^2\right|$$.

First, since $$z^2 + z + 1 = 0$$, $$z$$ is a primitive cube root of unity ($$z = \omega$$), so $$z^3 = 1$$ and $$|z| = 1$$, hence $$\frac{1}{z^n} = \bar{z}^n = z^{-n}$$.

Next, expanding the square gives $$\left(z^n + (-1)^n z^{-n}\right)^2 = z^{2n} + 2(-1)^n + z^{-2n}$$.

Now summing from $$n = 1$$ to $$15$$ yields $$ S = \sum_{n=1}^{15} z^{2n} + 2\sum_{n=1}^{15}(-1)^n + \sum_{n=1}^{15} z^{-2n}. $$

Since $$z^3 = 1$$, the terms $$z^{2n}$$ cycle every three values: $$z^2 + z^4 + z^6 + \cdots = (z^2 + z + 1) + (z^2 + z + 1) + \cdots$$. Each group sums to $$z^2 + z + 1 = 0$$ and with 15 terms (5 full cycles) one gets $$\sum_{n=1}^{15} z^{2n} = 0$$. Similarly, $$\sum_{n=1}^{15} z^{-2n} = 0$$.

For the alternating sum, one finds $$\sum_{n=1}^{15}(-1)^n = -1 + 1 - 1 + \cdots - 1 = -1$$ as there are 15 terms beginning with $$-1$$.

Therefore, $$ S = 0 + 2(-1) + 0 = -2, $$ so $$|S| = |-2| = 2$$.

Hence the answer is $$2$$.

We need all integers

$$n$$

satisfying

$$|2n^2-n-1|\le 800$$

Since

$$2n^2-n-1=(2n+1)(n-1)$$

and the parabola opens upward, solve

$$-800\le 2n^2-n-1\le 800$$

First,

$$2n^2-n-1\le 800$$

$$2n^2-n-801\le 0$$

The roots are

$$n=\frac{1\pm\sqrt{6409}}{4}=\frac{1\pm 80.056\ldots}{4}$$

Hence

$$-19.76\ldots \le n\le 20.26\ldots$$

So the integer values are

$$-19\le n\le 20$$

Now check the other inequality:

$$2n^2-n-1\ge -800$$

$$2n^2-n+799\ge 0$$

Its discriminant is

$$1-4(2)(799)=1-6392<0$$

and the quadratic opens upward, so it is always positive.

Therefore

$$S=\{-19,-18,\ldots,20\}$$

and

$$\sum_{n\in S}f(n)=\sum_{n=-19}^{20}(2n^2-n-1)$$

There are

$$40$$

terms.

Also,

$$\sum_{n=-19}^{20}n=\frac{40(-19+20)}{2}=20$$

Further,

$$\sum_{n=-19}^{20}n^2=\sum_{n=1}^{19}n^2+\sum_{n=1}^{20}n^2$$

$$=\frac{19\cdot20\cdot39}{6}+\frac{20\cdot21\cdot41}{6}$$

$$=2470+2870$$

$$=5340$$

Hence

$$\sum_{n\in S}f(n)=2(5340)-20-40$$

$$=10680-60$$

$$=10620$$

Final Answer :

$$10620$$

We need to find the number of Gaussian integers $$z = a + ib$$ (with $$a, b \in \mathbb{Z}$$) such that $$1 < |z - 3 + 2i| < 4$$.

This means $$1 < |(a-3) + i(b+2)| < 4$$, i.e.:

$$1 < \sqrt{(a-3)^2 + (b+2)^2} < 4$$

$$1 < (a-3)^2 + (b+2)^2 < 16$$

Let $$u = a - 3, v = b + 2$$. We need integer pairs $$(u, v)$$ with $$1 < u^2 + v^2 < 16$$, so $$u^2 + v^2 \in \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$$.

Total lattice points with $$u^2 + v^2 < 16$$: we enumerate for $$u^2 + v^2 \leq 15$$.

For each value of $$u^2 + v^2$$:

$$u^2+v^2 = 0$$: $$(0,0)$$ — 1 point

$$u^2+v^2 = 1$$: $$(\pm 1, 0), (0, \pm 1)$$ — 4 points

$$u^2+v^2 = 2$$: $$(\pm 1, \pm 1)$$ — 4 points

$$u^2+v^2 = 4$$: $$(\pm 2, 0), (0, \pm 2)$$ — 4 points

$$u^2+v^2 = 5$$: $$(\pm 1, \pm 2), (\pm 2, \pm 1)$$ — 8 points

$$u^2+v^2 = 8$$: $$(\pm 2, \pm 2)$$ — 4 points

$$u^2+v^2 = 9$$: $$(\pm 3, 0), (0, \pm 3)$$ — 4 points

$$u^2+v^2 = 10$$: $$(\pm 1, \pm 3), (\pm 3, \pm 1)$$ — 8 points

$$u^2+v^2 = 13$$: $$(\pm 2, \pm 3), (\pm 3, \pm 2)$$ — 8 points

Values $$u^2+v^2 = 3, 6, 7, 11, 12, 14, 15$$: no integer solutions (can be verified).

Total with $$u^2 + v^2 < 16$$: $$1 + 4 + 4 + 4 + 8 + 4 + 4 + 8 + 8 = 45$$

Exclude $$u^2 + v^2 \leq 1$$: $$1 + 4 = 5$$ points.

Answer: $$45 - 5 = 40$$

Hence the answer is $$\boxed{40}$$.

We have the complex number $$\sqrt{3}+i$$. First we express it in polar (modulus-argument) form, because raising a complex number to a high power is easiest when the number is written as $$r(\cos\theta+i\sin\theta)$$.

The modulus is found from the formula $$r=\sqrt{(\text{Re})^{2}+(\text{Im})^{2}}$$. Here $$\text{Re}=\sqrt{3}$$ and $$\text{Im}=1$$, so

$$r=\sqrt{(\sqrt{3})^{2}+1^{2}}=\sqrt{3+1}=2.$$

For the argument we use $$\theta=\tan^{-1}\left(\dfrac{\text{Im}}{\text{Re}}\right)$$. Hence

$$\theta=\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=\tan^{-1}(\tan 30^{\circ})=30^{\circ}=\dfrac{\pi}{6}.$$

So we can write

$$\sqrt{3}+i=2\bigl(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\bigr).$$

Now we are asked to evaluate $$(\sqrt{3}+i)^{100}$$. We will use De Moivre’s theorem, which states:

$$\bigl(r(\cos\theta+i\sin\theta)\bigr)^{n}=r^{\,n}\bigl(\cos n\theta+i\sin n\theta\bigr).$$

Applying the theorem with $$r=2,\;\theta=\dfrac{\pi}{6},\;n=100$$ gives

$$\bigl(\sqrt{3}+i\bigr)^{100}=2^{100}\bigl(\cos(100\cdot\tfrac{\pi}{6})+i\sin(100\cdot\tfrac{\pi}{6})\bigr).$$

The problem statement rewrites this power in the form $$2^{99}(p+iq)$$. To match that form we divide the expression we have by $$2^{99}$$:

$$\bigl(\sqrt{3}+i\bigr)^{100}=2^{99}\Bigl[2\bigl(\cos\tfrac{100\pi}{6}+i\sin\tfrac{100\pi}{6}\bigr)\Bigr]=2^{99}\bigl(p+iq\bigr).$$

Comparing the two brackets, we see directly that

$$p+iq=2\bigl(\cos\tfrac{100\pi}{6}+i\sin\tfrac{100\pi}{6}\bigr).$$

Now we simplify the angle. First compute

$$\dfrac{100\pi}{6}=\dfrac{50\pi}{3}.$$

Next, reduce $$\dfrac{50\pi}{3}$$ to an angle between $$0$$ and $$2\pi$$. Write

$$\dfrac{50\pi}{3}=16\pi+\dfrac{2\pi}{3}.$$

Because multiples of $$2\pi$$ do not change sine or cosine, we subtract $$16\pi=8(2\pi)$$ and keep only the remainder $$\dfrac{2\pi}{3}$$. Therefore

$$\cos\dfrac{50\pi}{3}=\cos\dfrac{2\pi}{3},\qquad\sin\dfrac{50\pi}{3}=\sin\dfrac{2\pi}{3}.$$

We know the exact trigonometric values:

$$\cos\dfrac{2\pi}{3}=-\dfrac12,\qquad\sin\dfrac{2\pi}{3}=\dfrac{\sqrt{3}}{2}.$$

Substituting these into the expression for $$p+iq$$ gives

$$p+iq=2\bigl(-\dfrac12+i\dfrac{\sqrt{3}}{2}\bigr)=-1+i\sqrt{3}.$$

Hence we read off

$$p=-1,\qquad q=\sqrt{3}.$$

We now form the quadratic equation whose roots are exactly $$p$$ and $$q$$. For a quadratic with roots $$\alpha$$ and $$\beta$$, the standard form is

$$x^{2}-(\alpha+\beta)x+\alpha\beta=0.$$

Here $$\alpha=p=-1$$ and $$\beta=q=\sqrt{3}$$, so

Sum of roots $$\alpha+\beta=(-1)+\sqrt{3}=\sqrt{3}-1,$$

Product of roots $$\alpha\beta=(-1)(\sqrt{3})=-\sqrt{3}.$$

Putting these into the standard form gives the required quadratic:

$$x^{2}-(\sqrt{3}-1)x-\sqrt{3}=0.$$

This matches Option D in the list given. Hence, the correct answer is Option 4.

We need to find the real part of $$(1 - \cos\theta + 2i\sin\theta)^{-1}$$ and set it equal to $$\frac{1}{5}$$.

Let $$z = 1 - \cos\theta + 2i\sin\theta$$. Then $$z^{-1} = \frac{\overline{z}}{|z|^2}$$.

The real part of $$z^{-1}$$ is $$\frac{\text{Re}(z)}{|z|^2} = \frac{1 - \cos\theta}{(1-\cos\theta)^2 + 4\sin^2\theta}$$.

We expand the denominator: $$(1-\cos\theta)^2 + 4\sin^2\theta = 1 - 2\cos\theta + \cos^2\theta + 4\sin^2\theta = 1 - 2\cos\theta + \cos^2\theta + 4(1 - \cos^2\theta)$$ $$= 5 - 2\cos\theta - 3\cos^2\theta.$$

Setting the real part equal to $$\frac{1}{5}$$: $$\frac{1-\cos\theta}{5 - 2\cos\theta - 3\cos^2\theta} = \frac{1}{5}.$$

Let $$c = \cos\theta$$. Then: $$5(1 - c) = 5 - 2c - 3c^2$$ $$5 - 5c = 5 - 2c - 3c^2$$ $$-5c = -2c - 3c^2$$ $$0 = 3c - 3c^2 = 3c(1 - c).$$

So $$c = 0$$ or $$c = 1$$. Since $$\theta \in (0, \pi)$$, we have $$\cos\theta \neq 1$$, so $$\cos\theta = 0$$, giving $$\theta = \frac{\pi}{2}$$.

Therefore, $$\int_0^\theta \sin x\, dx = \int_0^{\pi/2} \sin x\, dx = [-\cos x]_0^{\pi/2} = -\cos\frac{\pi}{2} + \cos 0 = 0 + 1 = 1.$$

We have $$w = 1 - \sqrt{3}\,i$$. First, $$|w| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$, and $$\arg(w) = -\frac{\pi}{3}$$ (since $$w$$ lies in the fourth quadrant with $$\tan^{-1}\frac{\sqrt{3}}{1} = \frac{\pi}{3}$$).

We need $$|zw| = 1$$, so $$|z| \cdot |w| = 1$$, giving $$|z| = \frac{1}{|w|} = \frac{1}{2}$$. Also $$\arg(z) - \arg(w) = \frac{\pi}{2}$$, so $$\arg(z) = \arg(w) + \frac{\pi}{2} = -\frac{\pi}{3} + \frac{\pi}{2} = \frac{\pi}{6}$$.

Therefore $$z = \frac{1}{2}\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = \frac{1}{2}\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = \frac{\sqrt{3}}{4} + \frac{i}{4}$$.

The area of the triangle with vertices at the origin $$O = (0,0)$$, $$z = \left(\frac{\sqrt{3}}{4}, \frac{1}{4}\right)$$, and $$w = (1, -\sqrt{3})$$ is given by $$\text{Area} = \frac{1}{2}|x_z \cdot y_w - x_w \cdot y_z|$$. Substituting: $$\text{Area} = \frac{1}{2}\left|\frac{\sqrt{3}}{4} \cdot (-\sqrt{3}) - 1 \cdot \frac{1}{4}\right| = \frac{1}{2}\left|-\frac{3}{4} - \frac{1}{4}\right| = \frac{1}{2} \times 1 = \frac{1}{2}$$.

The area of the triangle is $$\frac{1}{2}$$ square units, which is option (2).

We need to find the number of solutions of $$z^2 + 3\bar{z} = 0$$, where $$z$$ is a complex number.

Let $$z = x + iy$$, so $$\bar{z} = x - iy$$. The equation becomes $$(x + iy)^2 + 3(x - iy) = 0$$.

Expanding: $$x^2 + 2ixy - y^2 + 3x - 3iy = 0$$.

Separating real and imaginary parts: $$(x^2 - y^2 + 3x) + i(2xy - 3y) = 0$$.

For this to equal zero, both real and imaginary parts must be zero:

Real part: $$x^2 - y^2 + 3x = 0$$ $$-(1)$$

Imaginary part: $$2xy - 3y = 0$$ $$-(2)$$

From equation $$(2)$$: $$y(2x - 3) = 0$$, so either $$y = 0$$ or $$x = \frac{3}{2}$$.

Case 1: $$y = 0$$. Substituting into $$(1)$$: $$x^2 + 3x = 0$$, so $$x(x + 3) = 0$$, giving $$x = 0$$ or $$x = -3$$. This gives solutions $$z = 0$$ and $$z = -3$$.

Case 2: $$x = \frac{3}{2}$$. Substituting into $$(1)$$: $$\frac{9}{4} - y^2 + \frac{9}{2} = 0$$, so $$y^2 = \frac{9}{4} + \frac{9}{2} = \frac{27}{4}$$, giving $$y = \pm\frac{3\sqrt{3}}{2}$$. This gives two more solutions: $$z = \frac{3}{2} \pm \frac{3\sqrt{3}}{2}i$$.

So the total number of solutions is $$n = 4$$.

Now we compute $$\sum_{k=0}^{\infty} \frac{1}{n^k} = \sum_{k=0}^{\infty} \frac{1}{4^k} = \sum_{k=0}^{\infty} \left(\frac{1}{4}\right)^k$$.

This is an infinite geometric series with first term $$1$$ and common ratio $$r = \frac{1}{4}$$. Since $$|r| < 1$$, the sum is $$\frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$.

The answer is $$\frac{4}{3}$$, which is Option B.

Let $$z = x + iy$$. We need to find the intersection $$S_1 \cap S_2 \cap S_3$$.

For $$S_1$$: $$|z - 1| \leq \sqrt{2}$$ means $$(x-1)^2 + y^2 \leq 2$$. This is a closed disk centred at $$(1, 0)$$ with radius $$\sqrt{2}$$.

For $$S_2$$: $$\text{Re}((1-i)z) \geq 1$$. We compute $$(1-i)(x+iy) = x + iy - ix - i^2 y = (x+y) + i(y-x)$$. So $$\text{Re}((1-i)z) = x + y$$, and the condition becomes $$x + y \geq 1$$. This is the half-plane on and above the line $$x + y = 1$$.

For $$S_3$$: $$\text{Im}(z) \leq 1$$ means $$y \leq 1$$. This is the half-plane on and below the line $$y = 1$$.

Now we find the intersection. The line $$x + y = 1$$ intersects the circle $$(x-1)^2 + y^2 = 2$$. Substituting $$y = 1 - x$$ into the circle equation: $$(x-1)^2 + (1-x)^2 = 2$$, which gives $$2(x-1)^2 = 2$$, so $$(x-1)^2 = 1$$, meaning $$x = 0$$ or $$x = 2$$. The corresponding points are $$(0, 1)$$ and $$(2, -1)$$.

The line $$y = 1$$ intersects the circle: $$(x-1)^2 + 1 = 2$$, so $$(x-1)^2 = 1$$, giving $$x = 0$$ or $$x = 2$$. The points are $$(0, 1)$$ and $$(2, 1)$$.

The region $$S_1 \cap S_2 \cap S_3$$ is the part of the disk that lies above the line $$x + y = 1$$ and below the line $$y = 1$$. This is a closed region bounded by an arc of the circle from $$(0, 1)$$ to $$(2, -1)$$ (the upper arc), the line segment from $$(0, 1)$$ to $$(2, -1)$$ along $$x + y = 1$$, and capped at $$y = 1$$. Since the arc from $$(0,1)$$ to $$(2,1)$$ lies within $$y \leq 1$$ (the top of the circle is at $$y = \sqrt{2} \approx 1.414$$, which exceeds 1), the boundary also includes the line segment from $$(0,1)$$ to $$(2,1)$$ along $$y = 1$$.

This intersection region is a closed, bounded set with non-empty interior (it contains, for example, the point $$(1, 0.5)$$ which satisfies all three conditions). Since it has non-empty interior, it contains infinitely many elements.

The correct answer is Option C: has infinitely many elements.

Let us denote the required complex number by $$z$$ and write it in the usual Cartesian form

$$z = x + iy, \qquad \text{where } x, y \in \mathbb R.$$

The given condition is

$$\arg\!\left(\dfrac{z-1}{\,z+1\,}\right)=\dfrac{\pi}{4}.$$

For any complex number $$W = A + iB,$$ we recall the basic fact

$$\arg(W)=\theta \;\Longrightarrow\; \dfrac{B}{A}=\tan\theta.$$

Since $$\tan\!\left(\dfrac{\pi}{4}\right)=1,$$ the above gives the simple relation

$$B = A.$$

Thus, for the complex quotient

$$W=\dfrac{z-1}{z+1},$$

we merely have to equate its imaginary and real parts. We therefore proceed to calculate these two parts explicitly.

First write the numerator and the denominator in terms of $$x$$ and $$y$$:

$$z-1 = (x-1)+iy, \qquad z+1 = (x+1)+iy.$$

To find $$W$$ conveniently, multiply numerator and denominator by the complex conjugate of the denominator:

$$\dfrac{z-1}{z+1}= \dfrac{(x-1)+iy}{(x+1)+iy}\; \times\; \dfrac{(x+1)-iy}{(x+1)-iy}.$$

The denominator thus becomes the real number

$$\bigl(x+1\bigr)^2 + y^2.$$

Let us expand the numerator carefully, multiplying the two binomials term by term:

$$ \bigl[(x-1)+iy\bigr]\bigl[(x+1)-iy\bigr] = (x-1)(x+1) + (x-1)(-iy) + iy(x+1) + iy(-iy). $$

Compute each piece:

$$(x-1)(x+1) = x^2-1,$$
$$(x-1)(-iy) = -i y x + i y,$$
$$iy(x+1) = i y x + i y,$$
$$iy(-iy) = -i^2 y^2 = +y^2.$$

Adding the real parts and the imaginary parts separately we obtain

$$ x^2 - 1 + y^2 \;+\; 2 i y. $$

Hence

$$W=\dfrac{(x^2+y^2-1) + 2iy}{(x+1)^2 + y^2}.$$

From this expression we read off

$$\operatorname{Re}(W)=\dfrac{x^2 + y^2 -1}{(x+1)^2 + y^2}, \qquad \operatorname{Im}(W)=\dfrac{2y}{(x+1)^2 + y^2}.$$

Because the denominator is strictly positive, the equality of the imaginary and real parts entails

$$2y = x^2 + y^2 -1.$$

Let us bring every term to one side:

$$x^2 + y^2 - 2y - 1 = 0.$$

To recognise a circle we complete the square in the $$y$$-variable. Observe that

$$y^2 - 2y = (y-1)^2 - 1.$$

Substituting this into the equation gives

$$x^2 + \bigl[(y-1)^2 - 1\bigr] - 1 = 0.$$

Simplifying further,

$$x^2 + (y-1)^2 - 2 = 0,$$

or equivalently,

$$x^2 + (y-1)^2 = 2.$$

This is the standard form of a circle. Comparing with

$$ (x-h)^2 + (y-k)^2 = R^2,$$

we immediately read

Centre $$= (h,k) = (0,1), \qquad \text{Radius} \; R = \sqrt{2}.$$

Consequently the curve represented by the given argument condition is the circle centred at $$(0,1)$$ with radius $$\sqrt{2}$$.

Hence, the correct answer is Option D.

First, we simplify the right-hand side. We compute $$|5\sqrt{7} + 9i| = \sqrt{(5\sqrt{7})^2 + 9^2} = \sqrt{175 + 81} = \sqrt{256} = 16$$.

Then $$\log_{\sqrt{2}} 16 = \frac{\ln 16}{\ln \sqrt{2}} = \frac{4 \ln 2}{\frac{1}{2} \ln 2} = 8$$.

The inequality becomes: $$2^{\frac{(|z|+3)(|z|-1)}{|z|+1}} \geq 2^3$$ (since $$e^{a \ln 2} = 2^a$$ and $$||z|+1| = |z|+1$$ because $$|z| \geq 0$$).

Since the base 2 $$>$$ 1, we can compare exponents directly: $$\frac{(|z|+3)(|z|-1)}{|z|+1} \geq 3$$.

Let $$r = |z|$$ where $$r \geq 0$$. Multiplying both sides by $$(r+1)$$, which is always positive: $$(r+3)(r-1) \geq 3(r+1)$$.

Expanding: $$r^2 + 2r - 3 \geq 3r + 3$$, which simplifies to $$r^2 - r - 6 \geq 0$$.

Factoring: $$(r-3)(r+2) \geq 0$$. Since $$r \geq 0$$, the factor $$(r+2)$$ is always positive, so we need $$r - 3 \geq 0$$, giving $$r \geq 3$$.

The least value of $$|z|$$ is $$3$$.

Since $$\alpha, \beta \in \mathbb{R}$$, the polynomial $$z^2 + \alpha z + \beta = 0$$ has real coefficients. Because $$1 - 2i$$ is a root, its complex conjugate $$1 + 2i$$ must also be a root.

The sum of the roots gives $$-\alpha = (1 - 2i) + (1 + 2i) = 2$$, so $$\alpha = -2$$.

The product of the roots gives $$\beta = (1 - 2i)(1 + 2i) = 1 + 4 = 5$$.

Therefore, $$\alpha - \beta = -2 - 5 = -7$$.

We write the complex number in Cartesian form. Let $$z = x + iy$$ where $$x, y \in \mathbb R$$. Then

$$z - i \;=\; x + i(y - 1), \qquad z + 2i \;=\; x + i(y + 2).$$

We have the condition that the quotient $$\dfrac{z-i}{\,z+2i\,}$$ is a real number. A complex number is real exactly when its imaginary part is zero. So we require the imaginary part of the fraction to vanish.

To isolate the imaginary part, we multiply numerator and denominator by the conjugate of the denominator:

$$ \dfrac{z-i}{\,z+2i\,} = \dfrac{\,x + i(y-1)\,}{\,x + i(y+2)\,}\; = \dfrac{\,\bigl(x + i(y-1)\bigr)\,\bigl(x - i(y+2)\bigr)}{\,\bigl(x + i(y+2)\bigr)\,\bigl(x - i(y+2)\bigr)} = \dfrac{\,\bigl(x + i(y-1)\bigr)\,\bigl(x - i(y+2)\bigr)}{\,x^{2} + (y+2)^{2}}. $$

The denominator $$x^{2} + (y+2)^{2}$$ is real and positive (except at the point where it is zero, which we shall mention later). Therefore the entire fraction is real if and only if the numerator

$$\bigl(x + i(y-1)\bigr)\,\bigl(x - i(y+2)\bigr)$$

is itself a real number. Let us expand this product step by step. We recall the formula $$(a+ib)(c+id)=(ac-bd)+i(ad+bc).$$ Here we identify

$$a = x, \quad b = y-1, \quad c = x, \quad d = -(y+2).$$

First, the real part:

$$ac - bd \;=\; x\cdot x \;-\; (y-1)\bigl(-(y+2)\bigr) = x^{2} - \bigl(-(y^{2}+y-2)\bigr) = x^{2} + y^{2} + y - 2.$$

Next, the imaginary part:

$$ad + bc \;=\; x\bigl(-(y+2)\bigr) + (y-1)\,x = -x(y+2) + x(y-1) = x\bigl(-(y+2) + (y-1)\bigr) = x(-y-2+y-1) = -3x.$$

So the numerator equals

$$\bigl(x^{2} + y^{2} + y - 2\bigr) \;+\; i(-3x).$$

The imaginary part of the fraction is therefore $$-3x/(x^{2} + (y+2)^{2}).$$ For this to be zero we need

$$-3x = 0 \;\;\Longrightarrow\;\; x = 0.$$\

Thus every point in $$S$$ must satisfy $$x = 0$$, which is the equation of the imaginary axis. We must also ensure that the original denominator $$z + 2i$$ is not zero; this happens only when $$z = -2i$$. Hence

$$S = \bigl\{\,z = iy \;:\; y \in \mathbb R,\; y \neq -2\,\bigr\}.$$\

This set is a straight line (the imaginary axis) in the complex plane, missing a single point. A deleted point does not change the nature of the locus—it is still a straight line rather than a circle or a finite set.

Hence, the correct answer is Option D.

We need to solve $$\log_{\frac{1}{\sqrt{2}}}\left(\frac{|z| + 11}{(|z| - 1)^2}\right) \leq 2$$ where $$|z| \neq 1$$.

Let $$t = |z|$$ where $$t \geq 0$$ and $$t \neq 1$$. Since the base $$\frac{1}{\sqrt{2}}$$ is between 0 and 1, the logarithm is a decreasing function. Therefore, the inequality $$\log_{\frac{1}{\sqrt{2}}}(f(t)) \leq 2$$ becomes $$f(t) \geq \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$$.

So we need $$\frac{t + 11}{(t - 1)^2} \geq \frac{1}{2}$$.

This gives $$2(t + 11) \geq (t - 1)^2$$, i.e., $$2t + 22 \geq t^2 - 2t + 1$$.

Rearranging: $$0 \geq t^2 - 4t - 21$$, which means $$t^2 - 4t - 21 \leq 0$$.

Factoring: $$(t - 7)(t + 3) \leq 0$$.

This gives $$-3 \leq t \leq 7$$. Since $$t = |z| \geq 0$$ and $$t \neq 1$$, we have $$0 \leq t \leq 7$$ with $$t \neq 1$$.

We also need the argument of the logarithm to be positive, i.e., $$\frac{t + 11}{(t - 1)^2} > 0$$. Since $$t \geq 0$$, the numerator $$t + 11 > 0$$ always, and $$(t - 1)^2 > 0$$ for $$t \neq 1$$. So this is satisfied.

The largest value of $$|z|$$ is $$\mathbf{7}$$, which is option (2).

Let a general complex number be written as $$z = x + iy$$ where $$x = \text{Re}(z)$$ and $$y = \text{Im}(z)$$. We now translate the three set-descriptions into algebraic conditions in $$x$$ and $$y$$ so that we can see their mutual intersection clearly.

From $$S_1$$ we have

$$|z - (3 + 2i)|^2 = 8.$$

Using the fact that $$|a + ib|^2 = a^2 + b^2$$, we write

$$|(x+iy) - (3 + 2i)|^2 = |(x-3) + i(y-2)|^2 = (x-3)^2 + (y-2)^2.$$

So the condition for $$S_1$$ becomes

$$ (x-3)^2 + (y-2)^2 = 8. \quad -(1)$$

This is the equation of a circle with centre $$(3,\,2)$$ and radius $$\sqrt{8} = 2\sqrt2 \;(\approx 2.828).$$

From $$S_2$$ we are told that the real part is at least $$5$$, so

$$\text{Re}(z) \ge 5 \quad\Longrightarrow\quad x \ge 5. \quad -(2)$$

For $$S_3$$ we use the conjugate. Because $$\bar z = x - iy$$, we have

$$z - \bar z = (x + iy) - (x - iy) = 2iy.$$

Now, $$|2iy| = 2|y|,$$ therefore the condition $$|z - \bar z|\ge 8$$ gives

$$2|y| \ge 8 \quad\Longrightarrow\quad |y| \ge 4 \quad\Longrightarrow\quad y \ge 4 \text{ or } y \le -4. \quad -(3)$$

We now want the simultaneous satisfaction of (1), (2) and (3). In other words, we need all points on the circle (1) that also lie in the half-plane $$x\ge 5$$ and simultaneously have either $$y\ge 4$$ or $$y\le -4$$.

Because the circle’s centre is $$(3,2)$$ and its radius is $$2\sqrt2\approx 2.828,$$ its extreme $$x$$-coordinates are

$$x_{\min}=3-2\sqrt2\approx 0.172,\qquad x_{\max}=3+2\sqrt2\approx5.828.$$

Condition (2) restricts us to

$$5 \le x \le 5.828. \quad -(4)$$

Let us keep $$x$$ within the interval (4) and solve (1) for $$y$$. Starting from (1), we isolate the $$y$$-part:

$$ (y-2)^2 = 8 - (x-3)^2. $$

Taking the square root of both sides gives

$$ y - 2 = \pm\sqrt{\,8 - (x-3)^2\,}. \quad -(5)$$

Define

$$t = \sqrt{\,8 - (x-3)^2\,}, \qquad t\ge 0.$$

Then the two possible $$y$$ values are

$$ y = 2 + t \quad\text{or}\quad y = 2 - t. \quad -(6)$$

We next impose (3). First consider $$y \ge 4.$$ Using (6) we have

$$2 + t \ge 4 \quad\Longrightarrow\quad t \ge 2. \quad -(7)$$

Because $$t=\sqrt{\,8 - (x-3)^2\,},$$ inequality (7) squares to

$$8 - (x-3)^2 \ge 4 \quad\Longrightarrow\quad (x-3)^2 \le 4. \quad -(8)$$

Inequality (8) is the same as

$$-2 \le x - 3 \le 2 \quad\Longrightarrow\quad 1 \le x \le 5. \quad -(9)$$

Combining (9) with the earlier restriction (4), we must have

$$5 \le x \le 5.828\quad\text{and}\quad 1 \le x \le 5.$$

The only common value is

$$x = 5.$$

Substituting $$x=5$$ back into (1) to find $$y$$, we compute

$$(x-3)^2 = (5-3)^2 = 4,$$

and hence

$$(y-2)^2 = 8 - 4 = 4.$$

Therefore

$$y - 2 = \pm 2 \quad\Longrightarrow\quad y = 4 \ \text{or}\ y = 0.$$

From (3) we need $$|y| \ge 4,$$ so $$y = 0$$ is discarded while $$y = 4$$ is accepted.

We must also test the case $$y \le -4$$. However, the circle’s lowest $$y$$-coordinate is the centre coordinate minus the radius, namely $$2 - 2\sqrt2 \approx -0.828,$$ which is already greater than $$-4.$$ Thus no point on the circle can satisfy $$y\le -4.$$

Consequently, exactly one point survives all three conditions, namely

$$z = 5 + 4i.$$

There is only one such complex number, so the intersection $$S_1 \cap S_2 \cap S_3$$ contains exactly one element.

Hence, the correct answer is Option A.

Let us write the variable complex number as $$z = x + iy,\; x,y \in \mathbb{R}$$.

For the set $$S_1$$ we have the condition of a circle:

$$|z-2| \le 1 \; \Longrightarrow \; |(x+iy)-2| \le 1 \; \Longrightarrow\; (x-2)^2 + y^2 \le 1.$$

For the set $$S_2$$ start from the given expression and replace $$z$$ and $$\bar z$$:

$$z(1+i)+\bar z(1-i) \ge 4.$$

Because $$z = x+iy$$ and $$\bar z = x-iy,$$

$$z(1+i)= (x+iy)(1+i)=x(1+i)+iy(1+i) = (x - y) + i(x+y),$$

$$\bar z(1-i)= (x-iy)(1-i)=x(1-i)-iy(1-i) = (x - y) - i(x+y).$$

Adding both gives

$$\bigl[(x - y) + i(x+y)\bigr] + \bigl[(x - y) - i(x+y)\bigr] = 2(x-y).$$

Hence $$S_2$$ is the half-plane

$$2(x-y)\ge 4 \;\Longrightarrow\; x-y \ge 2 \;\Longrightarrow\; y \le x-2.$$

Therefore $$S_1\cap S_2$$ consists of the part of the closed disc $$(x-2)^2+y^2\le 1$$ that lies on or below the line $$y = x-2.$$

We have to maximise

$$\left|z-\frac52\right|^2 = (x-\tfrac52)^2 + y^2$$

subject to both constraints. The centre of the disc is $$C(2,0)$$ and the point from which we measure the distance is $$P\!\left(\frac52,0\right).$$ Since the distance is radially increasing, its maximum in the feasible region will occur on the boundary circle $$(x-2)^2+y^2 = 1.$$

Put $$x = 2 + \cos\theta,\; y = \sin\theta, \quad -\pi \le \theta \le \pi,$$ the standard parametrisation of this circle.

The half-plane condition $$y\le x-2$$ becomes

$$\sin\theta \le \cos\theta,$$

or

$$\cos\theta - \sin\theta \ge 0.$$

Using the identity $$\cos\theta - \sin\theta = \sqrt2\cos\!\left(\theta+\frac{\pi}{4}\right),$$ the inequality is $$\sqrt2\cos\!\left(\theta+\frac{\pi}{4}\right)\ge 0 \;\Longrightarrow\; -\frac{3\pi}{4}\le \theta \le \frac{\pi}{4}.$$

Now compute the required squared distance:

$$\left|z-\frac52\right|^2 = \bigl(2+\cos\theta-\tfrac52\bigr)^2 + (\sin\theta)^2 = (\cos\theta-\tfrac12)^2 + \sin^2\theta.$$

Expanding and using $$\sin^2\theta+\cos^2\theta=1,$$

$$\left|z-\frac52\right|^2 = \cos^2\theta - \cos\theta + \frac14 + \sin^2\theta = 1 - \cos\theta + \frac14 = \frac54 - \cos\theta.$$

Thus the problem reduces to maximising $$\frac54 - \cos\theta$$ over $$-\dfrac{3\pi}{4}\le\theta\le\dfrac{\pi}{4}.$$ Equivalently, we must minimise $$\cos\theta$$ on that interval. The cosine function attains its minimum at the left end-point $$\theta = -\dfrac{3\pi}{4},$$ where

$$\cos\!\left(-\frac{3\pi}{4}\right)= -\frac{\sqrt2}{2}.$$

Substituting this value,

$$\max\left|z-\frac52\right|^2 = \frac54 - \Bigl(-\frac{\sqrt2}{2}\Bigr) = \frac54 + \frac{\sqrt2}{2} = \frac{5 + 2\sqrt2}{4}.$$

Hence, the correct answer is Option D.

We need to find the equations of the two normal lines and the tangent line by converting from complex number form to Cartesian form. Let $$z = x + iy$$ and $$\bar{z} = x - iy$$.

For Line 1: $$(2 - i)z = (2 + i)\bar{z}$$. Expanding: $$(2 - i)(x + iy) = (2 + i)(x - iy)$$. The left side gives $$2x + 2iy - ix - i^2y = 2x + y + i(2y - x)$$. The right side gives $$2x - 2iy + ix - i^2y = 2x + y + i(x - 2y)$$. Equating imaginary parts: $$2y - x = x - 2y$$, which gives $$4y = 2x$$, or $$x = 2y$$.

For Line 2: $$(2 + i)z + (i - 2)\bar{z} - 4i = 0$$. Expanding: $$(2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0$$. This gives $$(2x - y + ix + 2iy) + (-2x + y + ix + 2iy) - 4i = 0$$, which simplifies to $$2ix + 4iy - 4i = 0$$. Dividing by $$2i$$: $$x + 2y - 2 = 0$$, or $$x + 2y = 2$$.

The centre of the circle is the intersection of the two normals. From $$x = 2y$$ and $$x + 2y = 2$$: substituting gives $$2y + 2y = 2$$, so $$y = \frac{1}{2}$$ and $$x = 1$$. The centre is $$\left(1, \frac{1}{2}\right)$$.

For the tangent line: $$iz + \bar{z} + 1 + i = 0$$. Expanding: $$i(x + iy) + (x - iy) + 1 + i = 0$$, which gives $$(ix - y) + (x - iy) + 1 + i = 0$$. Separating real and imaginary parts: $$(x - y + 1) + i(x - y + 1) = 0$$. Both parts give $$x - y + 1 = 0$$.

The radius equals the perpendicular distance from the centre $$\left(1, \frac{1}{2}\right)$$ to the tangent line $$x - y + 1 = 0$$: $$r = \frac{|1 - \frac{1}{2} + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{\frac{3}{2}}{\sqrt{2}} = \frac{3}{2\sqrt{2}}$$

Therefore, the radius of the circle is $$\dfrac{3}{2\sqrt{2}}$$, which corresponds to Option 3.

We are given three vertices in the complex plane: $$P(z)$$, $$Q(iz)$$, and $$R(z + iz)$$.

Let us observe the relationship between these points. The vector from $$P$$ to $$R$$ is $$R - P = (z + iz) - z = iz$$. The vector from $$Q$$ to $$R$$ is $$R - Q = (z + iz) - iz = z$$.

Notice that $$\overrightarrow{PR} = iz$$ and $$\overrightarrow{QR} = z$$. Multiplying $$z$$ by $$i$$ rotates it by $$90°$$ in the complex plane, so $$iz \perp z$$. This means $$PR \perp QR$$, and the triangle is right-angled at $$R$$.

The two sides meeting at the right angle have lengths $$|PR| = |iz| = |z|$$ and $$|QR| = |z|$$.

For a right triangle, the area is $$\frac{1}{2} \times \text{base} \times \text{height}$$. Here both legs equal $$|z|$$, so the area is $$\frac{1}{2} \times |z| \times |z| = \frac{1}{2}|z|^2$$.

This matches Option B: $$\frac{1}{2}|z|^2$$.

We are given $$|z\omega| = 1$$ and $$\arg(z) - \arg(\omega) = \frac{3\pi}{2}$$.

Let $$\arg(z) = \alpha$$ and $$\arg(\omega) = \beta$$. Since $$\bar{z}$$ has argument $$-\alpha$$, we get $$\arg(\bar{z}\omega) = -\alpha + \beta = -(\alpha - \beta) = -\frac{3\pi}{2}$$. Adding $$2\pi$$ to bring this into the principal range $$(-\pi, \pi]$$ gives $$\arg(\bar{z}\omega) = \frac{\pi}{2}$$. Also, $$|\bar{z}\omega| = |z\omega| = 1$$. Therefore $$\bar{z}\omega = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i$$.

Substituting $$\bar{z}\omega = i$$: $$\frac{1 - 2\bar{z}\omega}{1 + 3\bar{z}\omega} = \frac{1 - 2i}{1 + 3i}.$$

Multiplying numerator and denominator by the conjugate of the denominator: $$\frac{1 - 2i}{1 + 3i} \cdot \frac{1 - 3i}{1 - 3i} = \frac{(1)(1) + (1)(-3i) + (-2i)(1) + (-2i)(-3i)}{1^2 + 3^2} = \frac{1 - 3i - 2i + 6i^2}{10} = \frac{1 - 5i - 6}{10} = \frac{-5 - 5i}{10} = \frac{-1 - i}{2}.$$

The complex number $$w = \frac{-1-i}{2}$$ lies in the third quadrant (both real and imaginary parts are negative). Its reference angle is $$\arctan\!\left(\frac{1/2}{1/2}\right) = \frac{\pi}{4}$$. The principal argument (in $$(-\pi, \pi]$$) is therefore $$-\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$$.

Therefore $$\arg\!\left(\frac{1 - 2\bar{z}\omega}{1 + 3\bar{z}\omega}\right) = -\dfrac{3\pi}{4}$$.

Let us write the unknown complex number in its Cartesian form

$$z = x + iy,\qquad x,y \in \mathbb R.$$

The condition given in the problem is that the complex number

$$\dfrac{z-i}{\,z-1\,}$$

is purely imaginary, that is, its real part is zero. We therefore require

$$\operatorname{Re}\!\left(\dfrac{z-i}{z-1}\right)=0.$$

Substituting $$z = x + iy$$, we first write the numerator and the denominator separately:

$$z-i = (x+iy)-i = x + i(y-1),$$

$$z-1 = (x+iy)-1 = (x-1)+iy.$$

To extract the real part of a complex quotient we multiply numerator and denominator by the conjugate of the denominator. Using the identity

$$\dfrac{a+ib}{c+id} = \dfrac{(a+ib)(c-id)}{(c+id)(c-id)}$$

we get

$$\dfrac{z-i}{z-1}= \dfrac{\bigl[x+i(y-1)\bigr]\bigl[(x-1)-iy\bigr]}{(x-1)^2+y^2}.$$

The denominator $$(x-1)^2+y^2$$ is real and positive, so the real part of the whole fraction is obtained just from the real part of the numerator. We proceed to expand the numerator step by step:

$$$ \begin{aligned} \bigl[x+i(y-1)\bigr]\bigl[(x-1)-iy\bigr] &= x\,(x-1-iy)+i(y-1)\,(x-1-iy)\\[2mm] &= x(x-1)-ixy + i(y-1)(x-1) - i^2(y-1)y\\[2mm] &= x^2 - x - ixy + i(y-1)(x-1) + (y-1)y \end{aligned} $$$

Collecting the real and imaginary parts of the numerator:

Real part:

$$x^2 - x + y(y-1)=x^2 - x + y^2 - y,$$

Imaginary part:

$$-xy + (y-1)(x-1)= -xy + xy - x + y -1 = -x + y -1.$$

(The precise value of the imaginary part is unimportant for the present condition, but it is shown here for completeness.)

Because the overall denominator is real, the fraction is purely imaginary if and only if the real part of the numerator is zero. Therefore we must have

$$x^2 - x + y^2 - y = 0.$$

We now recognise this as the equation of a circle. Completing the square in both variables, we obtain

$$$ \begin{aligned} x^2 - x &= \left(x-\dfrac12\right)^2 - \dfrac14,\\[2mm] y^2 - y &= \left(y-\dfrac12\right)^2 - \dfrac14. \end{aligned} $$$

Hence

$$$ \left(x-\dfrac12\right)^2 - \dfrac14 + \left(y-\dfrac12\right)^2 - \dfrac14 = 0, $$$

which simplifies to

$$$ \left(x-\dfrac12\right)^2 + \left(y-\dfrac12\right)^2 = \dfrac12. $$$

Thus all admissible points $$z$$ lie on the circle with centre

$$C=\left(\dfrac12,\;\dfrac12\right)$$

and radius

$$r=\sqrt{\dfrac12}=\dfrac1{\sqrt2}.$$

Our goal is to minimise the distance from such a point $$z$$ to the fixed point

$$P = 3 + 3i \;\longrightarrow\; (3,3).$$

The distance between the centre $$C$$ and the point $$P$$ is

$$$ \begin{aligned} CP &= \sqrt{(3-\tfrac12)^2 + (3-\tfrac12)^2}\\[2mm] &= \sqrt{\left(\dfrac52\right)^2 + \left(\dfrac52\right)^2}\\[2mm] &= \dfrac52\sqrt{2}=2.5\sqrt2. \end{aligned} $$$

Because the point $$z$$ must stay on the circle of radius $$r$$, the smallest possible distance from $$P$$ to any such $$z$$ is obtained by moving directly towards $$P$$ from the centre $$C$$ and then stopping at the circle. Geometrically, this minimum distance is simply

$$\text{minimum distance}=CP - r.$$

Substituting the two values we have just found,

$$$ \begin{aligned} CP - r &= 2.5\sqrt2 - \dfrac1{\sqrt2}\\[2mm] &= \dfrac{2.5(\sqrt2)^2 - 1}{\sqrt2}\\[2mm] &= \dfrac{2.5\cdot 2 - 1}{\sqrt2}\\[2mm] &= \dfrac{5 - 1}{\sqrt2}\\[2mm] &= \dfrac4{\sqrt2}=2\sqrt2. \end{aligned} $$$

Therefore the minimum value of $$|z-(3+3i)|$$ is $$2\sqrt2$$.

Hence, the correct answer is Option B.

Let us denote the moving point by $$z=x+iy,\;x,y\in\mathbb R.$$

We are given the condition

$$\arg\!\left(\dfrac{z-2}{z+2}\right)=\dfrac{\pi}{4}.$$

For any complex number $$w=u+iv,$$ the argument equality $$\arg w=\dfrac{\pi}{4}$$ is equivalent to

$$\dfrac{\Im w}{\Re w}=\tan\dfrac{\pi}{4}=1 \quad\text{and}\quad \Re w>0.$$

So we must have

$$\Im\!\left(\dfrac{z-2}{z+2}\right)=\Re\!\left(\dfrac{z-2}{z+2}\right).$$

We now evaluate the real and imaginary parts of $$\dfrac{z-2}{z+2}.$$ Substituting $$z=x+iy$$, we get

$$\frac{z-2}{z+2}=\frac{(x-2)+iy}{(x+2)+iy}.$$

Multiply numerator and denominator by the conjugate of the denominator, $$(x+2)-iy,$$ to rationalise:

$$\frac{(x-2)+iy}{(x+2)+iy}\; \cdot\; \frac{(x+2)-iy}{(x+2)-iy} =\frac{[(x-2)(x+2)+y^{2}] + i[\,y(x+2)-y(x-2)\,]} {(x+2)^2+y^{2}}.$$

Calculating the two parts separately, we find

$$\begin{aligned} (x-2)(x+2)+y^{2}&=x^{2}-4+y^{2},\\ y(x+2)-y(x-2)&=y(x+2-x+2)=4y. \end{aligned}$$

Hence

$$\frac{z-2}{z+2} =\frac{\,x^{2}+y^{2}-4 + i\,4y\,}{x^{2}+y^{2}+4x+4}.$$

Therefore,

$$\Re\!\left(\frac{z-2}{z+2}\right)=\frac{x^{2}+y^{2}-4}{x^{2}+y^{2}+4x+4},\qquad \Im\!\left(\frac{z-2}{z+2}\right)=\frac{4y}{x^{2}+y^{2}+4x+4}.$$

Setting the imaginary part equal to the real part we obtain

$$\frac{4y}{x^{2}+y^{2}+4x+4} =\frac{x^{2}+y^{2}-4}{x^{2}+y^{2}+4x+4}.$$

The common denominator cancels out, leaving

$$4y=x^{2}+y^{2}-4.$$

Rearranging,

$$x^{2}+y^{2}-4y-4=0.$$

To recognise the geometrical locus, we complete the square in $$y$$:

$$y^{2}-4y=(y-2)^{2}-4,$$ so that

$$x^{2}+(y-2)^{2}-4-4=0 \;\Longrightarrow\; x^{2}+(y-2)^{2}=8.$$

Thus the point $$z$$ moves on the circle with centre $$C(0,2)$$ and radius

$$R=\sqrt{8}=2\sqrt2.$$

Next, we must minimise the squared distance

$$|\,z-9\sqrt2-2i\,|^{2}.$$

The fixed point whose distance from $$z$$ is being measured is

$$Q(9\sqrt2,\,2).$$

Notice that both the centre $$C(0,2)$$ of the circle and the point $$Q$$ have the same $$y$$-coordinate $$2$$. Hence the line $$y=2$$ is the line through both the centre and the fixed point. In such a situation, the closest point on the circle to $$Q$$ lies on the segment $$CQ$$ and is simply the point where that line meets the circle on the near side.

The distance between the centre and $$Q$$ is purely horizontal:

$$CQ=9\sqrt2-0=9\sqrt2.$$

For any circle, the minimum distance from an external point to the circle equals the distance from the point to the centre minus the radius. Using

$$\text{Minimum distance}=CQ-R,$$

we obtain

$$\text{Minimum distance}=9\sqrt2-2\sqrt2=7\sqrt2.$$

The question asks for the minimum value of the squared distance, so we square the above:

$$\bigl(7\sqrt2\bigr)^{2}=7^{2}\cdot2=49\cdot2=98.$$

So, the answer is $$98$$.

We have the set of complex numbers $$z$$ that satisfy $$|z-2-2i|\le 1$$. By definition of the modulus, this represents all the points whose distance from the fixed point $$2+2i$$ is at most $$1$$. In geometric language, it is the closed circle (disc) with centre $$C=2+2i$$ and radius $$r=1$$.

The expression whose maximum modulus we have to study is $$|3iz+6|$$. To simplify the notation, let us introduce a new complex variable $$w$$ given by

$$w=3iz+6.$$

This is a linear (affine) transformation of the complex plane. Such a map has two parts:

1. Multiplication by $$3i$$, which multiplies every length by $$|3i|=3$$ and rotates the plane by $$90^\circ$$, because multiplication by $$i$$ corresponds to a rotation of $$\frac{\pi}{2}$$ radians counter-clockwise.
2. Translation by $$6$$, which shifts every point $$6$$ units along the real axis.

Because linear maps take circles to circles (or possibly to lines when the radius is zero), the image of our original circle will again be a circle. We now compute its centre and radius one algebraic step at a time.

Let us write $$z$$ in the form $$z=(2+2i)+(z-(2+2i))$$. Substituting this into $$w$$ we obtain

$$\begin{aligned} w &=3i\Bigl[(2+2i)+(z-(2+2i))\Bigr]+6 \\ &=3i(2+2i)+3i\bigl(z-(2+2i)\bigr)+6. \end{aligned}$$

First we evaluate the constant term $$3i(2+2i)+6$$:

$$\begin{aligned} 3i(2+2i) &=3i\cdot2+3i\cdot2i\\ &=6i+6i^{2}\\ &=6i+6(-1)\\ &=-6+6i. \end{aligned}$$

Adding the real shift $$+6$$ gives

$$(-6+6i)+6 = 0+6i = 6i.$$

Hence the image circle has centre

$$O_w = 6i.$$

Next we look at the variable part $$3i\bigl(z-(2+2i)\bigr)$$. The factor $$3i$$ multiplies every distance by $$3$$, so the radius of the image circle is

$$R = 3\cdot r = 3\cdot1 = 3.$$

Therefore the set of all possible $$w$$ is the closed disc

$$|w-6i|\le 3,$$

i.e. a circle centred on the positive imaginary axis at height $$6$$, with radius $$3$$.

We are asked for the maximum possible value of $$|w|=|3iz+6|$$. Geometrically, $$|w|$$ is simply the distance of the point $$w$$ from the origin. Inside a fixed circle, the farthest point from the origin always lies on the boundary along the line that joins the origin to the circle’s centre. So we move from the centre $$6i$$ further straight outward by one radius.

The distance of the centre from the origin is

$$|6i|=6.$$

Adding one radius gives the maximal possible modulus:

$$|w|_{\max}=6+3=9.$$

The exact point on the circle where this maximum is reached is obtained by moving from $$6i$$ in the same (purely imaginary) direction by length $$3$$:

$$w_{\max}=6i+3i=9i.$$

Thus the extremal value occurs at the complex number

$$w_{\max}=9i.$$

To find the corresponding $$z$$ we solve the linear equation $$w=3iz+6$$ with $$w=9i$$:

$$\begin{aligned} 3iz+6 &=9i\\ 3iz &=9i-6\\ &=-6+9i. \end{aligned}$$

Now we divide by $$3i$$. The reciprocal of $$i$$ is $$-i$$ because $$i\cdot(-i)=1$$. Proceeding step by step,

$$\begin{aligned} z &=\frac{-6+9i}{3i}\\[4pt] &=\frac{3(-2+3i)}{3i}\\[4pt] &=-\frac{2-3i}{i}\\[4pt] &=(-2+3i)\cdot(-i)\quad(\text{since }1/i=-i)\\[4pt] &=2i+3(-i^2)\\[4pt] &=2i+3(1)\\[4pt] &=3+2i. \end{aligned}$$

Thus the extremal point in the original $$z$$-plane is

$$z_{\max}=3+2i.$$

Comparing with the statement of the problem, this is $$a+ib$$ with $$a=3$$ and $$b=2$$. Therefore

$$a+b = 3+2 = 5.$$

So, the answer is $$5$$.

Let

$$z=x+iy$$

Then,

$$x+iy+\alpha\sqrt{(x-1)^2+y^2}+2i=0$$

Equating real and imaginary parts,

$$x+\alpha\sqrt{(x-1)^2+y^2}=0 \quad\cdots(1)$$ and $$y+2=0$$

Hence,

$$y=-2$$

Substitute into (1):

$$x+\alpha\sqrt{(x-1)^2+4}=0$$

$$\alpha=-\frac{x}{\sqrt{(x-1)^2+4}}$$

Therefore,

$$\alpha^2=\frac{x^2}{(x-1)^2+4}$$

$$=\frac{x^2}{x^2-2x+5}$$

Let

$$f(x)=\frac{x^2}{x^2-2x+5}$$

Differentiate:

$$f'(x)=\frac{2x(x^2-2x+5)-x^2(2x-2)}{(x^2-2x+5)^2}$$

$$=\frac{2x(5-x)}{(x^2-2x+5)^2}$$

Hence critical points are $$x=0,\qquad x=5$$

Now,

$$f(0)=0$$

and

$$f(5)=\frac{25}{20}$$

$$=\frac54$$

Thus,

$$0\le\alpha^2\le\frac54$$

Hence,

$$-\frac{\sqrt5}{2}\le\alpha\le\frac{\sqrt5}{2}$$

Therefore,

$$p=-\frac{\sqrt5}{2},\qquad q=\frac{\sqrt5}{2}$$

Now,

$$4(p^2+q^2)=4\left(\frac54+\frac54\right)$$

$$=4\cdot\frac52$$

$$=10$$

Therefore, the required value is

$$\boxed{10}$$

Given $$\left|\frac{z+i}{z-3i}\right| = 1$$, this means the point $$z$$ is equidistant from $$-i$$ and $$3i$$ in the complex plane. Setting $$z = x + iy$$, we get $$|z + i|^2 = |z - 3i|^2$$, which gives $$x^2 + (y+1)^2 = x^2 + (y-3)^2$$. Expanding: $$y^2 + 2y + 1 = y^2 - 6y + 9$$, so $$8y = 8$$, hence $$y = 1$$. Therefore $$z = x + i$$ for real $$x$$.

Now $$w = z\bar{z} - 2z + 2$$. With $$z = x + i$$, we have $$z\bar{z} = |z|^2 = x^2 + 1$$. So $$w = x^2 + 1 - 2(x + i) + 2 = x^2 - 2x + 3 - 2i$$.

The real part of $$w$$ is $$\text{Re}(w) = x^2 - 2x + 3 = (x-1)^2 + 2$$. This is minimized when $$x = 1$$, giving $$\text{Re}(w) = 2$$.

At $$x = 1$$: $$w = 2 - 2i$$. We can write $$w = 2 - 2i = 2\sqrt{2}\,e^{-i\pi/4}$$.

Then $$w^n = (2\sqrt{2})^n \, e^{-in\pi/4}$$. For $$w^n$$ to be real, we need $$e^{-in\pi/4}$$ to be real, which requires $$\frac{n\pi}{4} = k\pi$$ for some integer $$k$$, i.e., $$n = 4k$$.

The minimum natural number $$n$$ is $$n = 4$$.

Let us denote a general complex number by $$z=x+iy$$ where $$x=\text{Re}(z)$$ and $$y=\text{Im}(z)$$.

Both $$z_1$$ and $$z_2$$ satisfy the given locus

$$|z-3|=\text{Re}(z).$$

First, we recall the modulus formula: for any $$z=x+iy,$$ we have $$|z|=\sqrt{x^{2}+y^{2}}.$$ Applying this to $$z-3=(x-3)+iy$$, we get

$$|(x-3)+iy|=\sqrt{(x-3)^2+y^2}.$$

The given equation therefore becomes

$$\sqrt{(x-3)^2+y^2}=x.$$

Squaring both sides (because both sides are non-negative) gives

$$(x-3)^2+y^2=x^2.$$

Expanding the left side and cancelling $$x^2$$ from both sides, we obtain

$$x^2-6x+9+y^2=x^2 \;\;\Longrightarrow\;\; -6x+9+y^2=0.$$

Rearranging,

$$y^2=6x-9.$$

Thus every point of the locus must satisfy

$$y^2=6x-9,\qquad x\ge \tfrac32.$$

Next we use the second condition. Because

$$\arg(z_1-z_2)=\frac{\pi}{4},$$

the vector $$z_1-z_2$$ makes an angle of $$45^{\circ}$$ with the positive real axis. Therefore its slope is

$$\tan\!\frac{\pi}{4}=1,$$

so we must have

$$\frac{y_1-y_2}{x_1-x_2}=1 \quad\Longrightarrow\quad y_1-y_2=x_1-x_2.$$

This equality tells us that both points lie on the same straight line of slope $$1$$. Writing the equation of that line as

$$y=x+c$$

for some real constant $$c,$$ we know that each of $$z_1$$ and $$z_2$$ satisfies this equation together with the parabola equation derived earlier.

We therefore substitute $$y=x+c$$ into $$y^2=6x-9$$:

$$(x+c)^2=6x-9.$$

Expanding and collecting like terms gives the quadratic in $$x$$

$$x^2+2cx+c^2-6x+9=0,$$

or

$$x^2+(2c-6)x+(c^2+9)=0.$$

Let $$x_1$$ and $$x_2$$ be the roots of this quadratic; they are precisely the real parts of $$z_1$$ and $$z_2$$. By Vieta’s formulas, for a quadratic $$x^2+px+q=0,$$ the sum of the roots is $$-p$$ and the product is $$q$$. Hence here

$$x_1+x_2=-(2c-6)=6-2c.$$

Because $$y=x+c,$$ the corresponding imaginary parts are

$$y_1=x_1+c,\qquad y_2=x_2+c.$$

Adding these two imaginary parts, we get

$$y_1+y_2=(x_1+x_2)+2c=(6-2c)+2c=6.$$

The quantity asked for in the question is the imaginary part of $$z_1+z_2$$, and that is exactly $$y_1+y_2$$. We therefore have

$$\text{Im}(z_1+z_2)=6.$$

So, the answer is $$6$$.

We are given the complex number $$z=\dfrac{3+2i\cos\theta}{1-3i\cos\theta}$$ with $$\theta\in\left(0,\dfrac{\pi}{2}\right)$$ and the information that the real part of $$z$$ is zero. Our goal is to find the value of $$\sin^2 3\theta+\cos^2\theta$$.

First, we rewrite $$z$$ in the standard (a + ib) form so that its real part becomes evident. To do this, we multiply the numerator and the denominator by the complex conjugate of the denominator:

$$z=\dfrac{3+2i\cos\theta}{1-3i\cos\theta}\cdot\dfrac{1+3i\cos\theta}{1+3i\cos\theta} =\dfrac{(3+2i\cos\theta)(1+3i\cos\theta)}{(1)^2-(3i\cos\theta)^2}.$$

We now simplify the denominator. Using $$i^2=-1$$, we have

$$ (1)^2-(3i\cos\theta)^2 =1-\bigl(9i^2\cos^2\theta\bigr) =1-9(-1)\cos^2\theta =1+9\cos^2\theta. $$

Next we expand the numerator term by term:

$$ (3+2i\cos\theta)(1+3i\cos\theta) =3(1)+3(3i\cos\theta)+2i\cos\theta(1)+2i\cos\theta(3i\cos\theta). $$

Evaluating each product:

$$\begin{aligned} 3(1) &= 3,\\ 3(3i\cos\theta) &= 9i\cos\theta,\\ 2i\cos\theta(1) &= 2i\cos\theta,\\ 2i\cos\theta(3i\cos\theta) &= 6i^2\cos^2\theta =6(-1)\cos^2\theta =-6\cos^2\theta. \end{aligned}$$

Adding the real parts and the imaginary parts separately, we obtain

$$ (3-6\cos^2\theta) \;+\; i(9\cos\theta+2\cos\theta) =(3-6\cos^2\theta)\;+\;11i\cos\theta. $$

Therefore,

$$ z=\dfrac{(3-6\cos^2\theta)+11i\cos\theta}{1+9\cos^2\theta}. $$

In this form the real part of $$z$$ is clearly

$$\operatorname{Re}(z)=\dfrac{3-6\cos^2\theta}{1+9\cos^2\theta}.$$

Because the real part is zero, its numerator must be zero:

$$3-6\cos^2\theta=0.$$ Dividing by 3, we get $$1-2\cos^2\theta=0,$$ so $$\cos^2\theta=\dfrac{1}{2}.$$

Given that $$\theta\in\left(0,\dfrac{\pi}{2}\right)$$ (the first quadrant), we take the positive root:

$$\cos\theta=\dfrac{1}{\sqrt2}\quad\Longrightarrow\quad\theta=\dfrac{\pi}{4}.$$

Now we turn to the required expression $$\sin^2 3\theta+\cos^2\theta$$. We already know $$\cos^2\theta=\dfrac12$$. We must compute $$\sin 3\theta$$.

We recall the triple‐angle identity for sine: $$\sin 3A = 3\sin A - 4\sin^3 A.$$

Putting $$A=\theta=\dfrac{\pi}{4}$$, we first note $$\sin\theta=\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}.$$

Applying the formula,

$$ \begin{aligned} \sin 3\theta &=3\sin\theta-4\sin^3\theta\\ &=3\left(\dfrac{1}{\sqrt2}\right)-4\left(\dfrac{1}{\sqrt2}\right)^3\\ &=\dfrac{3}{\sqrt2}-4\cdot\dfrac{1}{2\sqrt2}\\ &=\dfrac{3}{\sqrt2}-\dfrac{2}{\sqrt2}\\ &=\dfrac{1}{\sqrt2}. \end{aligned} $$

Consequently,

$$\sin^2 3\theta=\left(\dfrac{1}{\sqrt2}\right)^2=\dfrac12.$$

Finally, we add the two squares:

$$ \sin^2 3\theta+\cos^2\theta =\dfrac12+\dfrac12 =1. $$

Hence, the correct answer is Option 1.

We first convert each complex number to polar form. We have $$-1 + i\sqrt{3} = 2e^{i \cdot 2\pi/3}$$, so $$(-1 + i\sqrt{3})^{21} = 2^{21} e^{i \cdot 42\pi/3} = 2^{21} e^{i \cdot 14\pi} = 2^{21}$$.

Also, $$1 - i = \sqrt{2}\, e^{-i\pi/4}$$, so $$(1 - i)^{24} = (\sqrt{2})^{24} e^{-i \cdot 6\pi} = 2^{12} \cdot 1 = 2^{12}$$.

Therefore, the first fraction is $$\frac{2^{21}}{2^{12}} = 2^9 = 512$$.

Similarly, $$1 + i\sqrt{3} = 2e^{i\pi/3}$$, so $$(1 + i\sqrt{3})^{21} = 2^{21} e^{i \cdot 7\pi} = 2^{21} e^{i\pi} = -2^{21}$$.

And $$1 + i = \sqrt{2}\, e^{i\pi/4}$$, so $$(1 + i)^{24} = 2^{12} e^{i \cdot 6\pi} = 2^{12}$$.

Therefore, the second fraction is $$\frac{-2^{21}}{2^{12}} = -2^9 = -512$$.

Adding the two fractions: $$k = 512 + (-512) = 0$$, so $$|k| = 0$$ and $$n = [|k|] = 0$$.

Now we compute $$\sum_{j=0}^{n+5}(j+5)^2 - \sum_{j=0}^{n+5}(j+5) = \sum_{j=0}^{5}(j+5)^2 - \sum_{j=0}^{5}(j+5)$$. Substituting $$m = j + 5$$, this becomes $$\sum_{m=5}^{10} m^2 - \sum_{m=5}^{10} m = (25 + 36 + 49 + 64 + 81 + 100) - (5 + 6 + 7 + 8 + 9 + 10) = 355 - 45 = 310$$.

The correct answer is $$310$$.

Let $$z = x + iy$$. The condition $$|z + 5| \leq 4$$ represents a closed disk centered at $$(-5, 0)$$ with radius 4, i.e., $$(x + 5)^2 + y^2 \leq 16$$.

For the second condition: $$z(1 + i) + \bar{z}(1 - i) = (x + iy)(1 + i) + (x - iy)(1 - i) = 2x - 2y$$. So the condition becomes $$2x - 2y \geq -10$$, or $$x - y \geq -5$$.

We need to maximize $$|z + 1|^2 = (x + 1)^2 + y^2$$.

The distance from the center $$(-5, 0)$$ of the disk to the line $$x - y + 5 = 0$$ is $$\frac{|-5 - 0 + 5|}{\sqrt{2}} = 0$$. So the line passes through the center of the disk, and the feasible region is a semicircular disk.

Parametrize the boundary circle as $$x = -5 + 4\cos\theta$$, $$y = 4\sin\theta$$. The condition $$x - y \geq -5$$ becomes $$4\cos\theta - 4\sin\theta \geq 0$$, i.e., $$\cos\theta \geq \sin\theta$$, which gives $$\theta \in [-\frac{3\pi}{4}, \frac{\pi}{4}]$$.

Now $$|z + 1|^2 = (x + 1)^2 + y^2 = (-4 + 4\cos\theta)^2 + 16\sin^2\theta = 16 - 32\cos\theta + 16\cos^2\theta + 16\sin^2\theta = 32 - 32\cos\theta$$.

This is maximized when $$\cos\theta$$ is minimized. Over $$\theta \in [-\frac{3\pi}{4}, \frac{\pi}{4}]$$, the minimum of $$\cos\theta$$ occurs at $$\theta = -\frac{3\pi}{4}$$, where $$\cos\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$.

Therefore, the maximum value of $$|z + 1|^2 = 32 - 32\left(-\frac{\sqrt{2}}{2}\right) = 32 + 16\sqrt{2}$$.

So $$\alpha = 32$$ and $$\beta = 16$$, giving $$\alpha + \beta = 48$$.

We have $$z=\dfrac{1-i\sqrt 3}{2}$$ and $$i=\sqrt{-1}\,.$$

First we find the modulus and argument of $$z$$. The modulus is

$$$|z|=\sqrt{\left(\dfrac12\right)^2+\left(\dfrac{-\sqrt3}{2}\right)^2}= \sqrt{\dfrac14+\dfrac34}= \sqrt1 =1.$$$

Because $$|z|=1$$, the number lies on the unit circle. Writing $$z=x+iy$$, we have $$x=\dfrac12$$ and $$y=-\dfrac{\sqrt3}{2}$$, so

$$\cos\theta=\dfrac12,\qquad\sin\theta=-\dfrac{\sqrt3}{2}.$$

These are the cosine and sine of $$-\dfrac{\pi}{3}$$. Hence

$$z=e^{-i\pi/3}.$$

For every positive integer $$k$$ we therefore get

$$$z^{\,k}=e^{-ik\pi/3},\qquad \dfrac1{z^{\,k}}=e^{ik\pi/3}.$$$

Adding these two expressions gives the well-known Euler relation

$$$z^{\,k}+\dfrac1{z^{\,k}}=e^{-ik\pi/3}+e^{ik\pi/3}=2\cos\!\left(\dfrac{k\pi}{3}\right).$$$

Now we cube both sides:

$$$\left(z^{\,k}+\dfrac1{z^{\,k}}\right)^3=\left(2\cos\!\left(\dfrac{k\pi}{3}\right)\right)^3 =8\cos^3\!\left(\dfrac{k\pi}{3}\right).$$$

The expression asked in the question is

$$$S=21+\sum_{k=1}^{21}\left(z^{\,k}+\dfrac1{z^{\,k}}\right)^3 =21+\sum_{k=1}^{21}8\cos^3\!\left(\dfrac{k\pi}{3}\right).$$$

Taking the constant $$8$$ outside the summation we get

$$S=21+8\sum_{k=1}^{21}\cos^3\!\left(\dfrac{k\pi}{3}\right).$$

To evaluate the trigonometric sum, we note that $$\cos\theta$$ has period $$2\pi$$, so $$\cos^3\theta$$ is also periodic with period $$2\pi$$. An increment of $$\theta$$ by $$\dfrac{\pi}{3}$$ corresponds to increasing $$k$$ by $$1$$. Hence every six successive values of $$k$$ produce one complete period:

$$$k=0,1,2,3,4,5 \quad\Longrightarrow\quad \theta=0,\dfrac{\pi}{3},\dfrac{2\pi}{3},\pi,\dfrac{4\pi}{3},\dfrac{5\pi}{3}.$$$

We list the individual cube values within one period:

$$$ \begin{aligned} k=0:&\;\cos^3 0 = 1,\\ k=1:&\;\cos^3\!\left(\dfrac{\pi}{3}\right)=\left(\dfrac12\right)^3=\dfrac18,\\ k=2:&\;\cos^3\!\left(\dfrac{2\pi}{3}\right)=\left(-\dfrac12\right)^3=-\dfrac18,\\ k=3:&\;\cos^3\pi = (-1)^3=-1,\\ k=4:&\;\cos^3\!\left(\dfrac{4\pi}{3}\right)=\left(-\dfrac12\right)^3=-\dfrac18,\\ k=5:&\;\cos^3\!\left(\dfrac{5\pi}{3}\right)=\left(\dfrac12\right)^3=\dfrac18. \end{aligned} $$$

Adding these six numbers gives

$$1+\dfrac18-\dfrac18-1-\dfrac18+\dfrac18 = 0.$$

Thus the sum of $$\cos^3\!\left(\dfrac{k\pi}{3}\right)$$ over any block of six consecutive integers $$k$$ is zero.

From $$k=1$$ to $$k=21$$ we have three full blocks of six terms (for $$k=1\text{ to }18$$) and then three extra terms (for $$k=19,20,21$$).

The three complete blocks contribute

$$3\times 0 = 0.$$

We now calculate the remaining three terms explicitly:

$$$ \begin{aligned} k=19:&\;19\bmod 6=1\;\Rightarrow\;\cos^3\!\left(\dfrac{19\pi}{3}\right)=\cos^3\!\left(\dfrac{\pi}{3}\right)=\dfrac18,\\ k=20:&\;20\bmod 6=2\;\Rightarrow\;\cos^3\!\left(\dfrac{20\pi}{3}\right)=\cos^3\!\left(\dfrac{2\pi}{3}\right)=-\dfrac18,\\ k=21:&\;21\bmod 6=3\;\Rightarrow\;\cos^3\!\left(\dfrac{21\pi}{3}\right)=\cos^3\pi=-1. \end{aligned} $$$

Adding these residual values gives

$$\dfrac18-\dfrac18-1=-1.$$

Therefore the complete trigonometric sum is

$$\sum_{k=1}^{21}\cos^3\!\left(\dfrac{k\pi}{3}\right) = -1.$$

Substituting back into $$S$$ we obtain

$$$S = 21 + 8(-1) = 21 - 8 = 13.$$$

So, the answer is $$13$$.

Let us begin by translating the given complex-number condition into an equation in the real variables $$x$$ and $$y$$.

We write $$z=x+iy$$, so that

$$z^{2}=(x+iy)^{2}=x^{2}-y^{2}+2ixy.$$

The real part of $$z^{2}$$ is therefore

$$\text{Re}(z^{2})=x^{2}-y^{2}.$$

Also, by definition, $$\text{Im}(z)=y,$$ whence

$$(\text{Im}(z))^{2}=y^{2}.$$

The given equation of the circle is

$$\text{Re}(z^{2})+2(\text{Im}(z))^{2}+2\text{Re}(z)=0.$$

Substituting the expressions we have just obtained, we get

$$\bigl(x^{2}-y^{2}\bigr)+2y^{2}+2x=0.$$

Combining like terms in $$y^{2}$$ leads to

$$x^{2}-y^{2}+2y^{2}+2x=0 \;\Longrightarrow\; x^{2}+y^{2}+2x=0.$$

To recognise the centre and radius of the circle, we complete the square in $$x$$:

$$x^{2}+2x=(x+1)^{2}-1.$$

Hence the equation becomes

$$(x+1)^{2}-1+y^{2}=0 \;\Longrightarrow\; (x+1)^{2}+y^{2}=1.$$

We now see that the circle has centre $$(-1,0)$$ and radius $$1$$.

Next, consider the parabola whose equation is

$$x^{2}-6x-y+13=0.$$

We rewrite it in the more familiar form $$y=x^{2}-6x+13.$$ For a quadratic $$y=ax^{2}+bx+c,$$ the vertex has $$x\text{-coordinate}=-\dfrac{b}{2a}.$$

Here $$a=1$$ and $$b=-6,$$ so

$$x_{v}=-\dfrac{-6}{2\cdot1}=3.$$

Substituting $$x=3$$ back into the expression for $$y$$ gives

$$y_{v}=3^{2}-6\cdot3+13=9-18+13=4.$$

Thus, the vertex of the parabola is $$\bigl(3,4\bigr).$$

We now require the straight line that passes through the two points $$(-1,0)$$ (the circle’s centre) and $$\bigl(3,4\bigr)$$ (the parabola’s vertex).

The slope (gradient) of a line through $$(x_{1},y_{1})$$ and $$(x_{2},y_{2})$$ is given by the formula $$m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}.$$

Applying this formula, we have

$$m=\dfrac{4-0}{3-(-1)}=\dfrac{4}{4}=1.$$

With slope $$m=1$$ and passing through, say, the point $$(3,4),$$ the point-slope form $$y-y_{1}=m(x-x_{1})$$ yields

$$y-4=1\bigl(x-3\bigr).$$

Simplifying, we find the equation of the line:

$$y=x+1.$$

The $$y$$-intercept of a line is its $$y$$-value when $$x=0.$$ Setting $$x=0$$ above gives

$$y=0+1=1.$$

Therefore, the required $$y$$-intercept equals $$1.$$

So, the answer is $$1$$.

We have to find the smallest positive integer $$n$$ for which the complex-valued expression

$$\dfrac{(2i)^n}{(1-i)^{\,n-2}}$$

comes out to be a positive integer (a real number > 0).

First we separate magnitude and argument of every factor.

The numerator is $$ (2i)^n = 2^n i^n $$. Here $$i = e^{\,i\pi/2}$$, so
$$ i^n = e^{\,i n\pi/2}. $$ Hence

$$ (2i)^n = 2^n\,e^{\,i n\pi/2}. $$

Now we treat the denominator. Write $$1-i$$ in polar form:

$$ 1-i = \sqrt2\left(\cos\!\bigl(-\tfrac{\pi}{4}\bigr)+i\sin\!\bigl(-\tfrac{\pi}{4}\bigr)\right)=\sqrt2\,e^{-\,i\pi/4}. $$

Therefore

$$ (1-i)^{\,n-2}= \bigl(\sqrt2\bigr)^{\,n-2}\,e^{-\,i(n-2)\pi/4}. $$

So the entire fraction becomes

$$ \dfrac{(2i)^n}{(1-i)^{\,n-2}} =\dfrac{2^n\,e^{\,i n\pi/2}}{\bigl(\sqrt2\bigr)^{\,n-2}\,e^{-\,i (n-2)\pi/4}} =2^n\bigl(\sqrt2\bigr)^{-(n-2)}\,e^{\,i\left(\dfrac{n\pi}{2}+\dfrac{(n-2)\pi}{4}\right)}. $$

We now simplify the magnitude (modulus).

Formula used: $$\bigl(\sqrt2\bigr)^{n-2}=2^{\,(n-2)/2}.$$ Substituting:

$$ 2^n \times 2^{-\,(n-2)/2}=2^{\,n-\frac{n-2}{2}}=2^{\,n-\frac{n}{2}+1}=2^{\,\frac{n}{2}+1}. $$

Thus the modulus of the fraction is $$2^{\,\frac{n}{2}+1}$$, clearly a positive real number for every positive integer $$n$$.

Next we handle the argument (angle). Adding the angles we obtained earlier, we have

$$ \text{Arg} = \dfrac{n\pi}{2} + \dfrac{(n-2)\pi}{4} = \dfrac{2n\pi + (n-2)\pi}{4} = \dfrac{(3n-2)\pi}{4}. $$

For the overall complex number to be a positive real, its argument must be an integral multiple of $$2\pi$$:

$$ \dfrac{(3n-2)\pi}{4}=2k\pi \quad\Longrightarrow\quad \dfrac{3n-2}{4}=2k \quad\Longrightarrow\quad 3n-2=8k, $$

where $$k$$ is any integer.

Rearranging,

$$ 3n = 8k+2. $$

We want the least positive $$n$$ that satisfies this linear Diophantine equation. Work modulo 3: $$8k+2\equiv (-1)k+2\pmod 3,$$ so we require

$$ (-1)k+2\equiv 0\pmod 3 \;\;\Longrightarrow\;\; k\equiv 2\pmod 3. $$

Put $$k=3m+2$$ with integer $$m\ge 0$$. Substituting back:

$$ 3n = 8(3m+2)+2 = 24m+16+2 = 24m+18 = 6(4m+3). $$

Hence

$$ n = 2(4m+3) = 8m+6. $$

The smallest value arises when $$m=0$$, giving

$$ n = 6. $$

For this $$n$$ the modulus is $$2^{\,\frac{6}{2}+1}=2^{\,4}=16$$, and the argument is zero, so the fraction equals $$16$$, a positive integer exactly as required.

Hence, the correct answer is Option 6.

We begin with the quadratic equation $$x^{2}+bx+45=0$$, where $$b\in\mathbb R$$. Its two roots are stated to be conjugate complex numbers, so we may write them as

$$z=\alpha+i\beta\quad\text{and}\quad\bar z=\alpha-i\beta,\qquad\beta\neq0.$$

For a quadratic $$ax^{2}+bx+c=0$$ with roots $$r_{1},r_{2}$$, Vieta’s relations give

$$r_{1}+r_{2}=-\dfrac{b}{a},\qquad r_{1}r_{2}=\dfrac{c}{a}.$$

In our case $$a=1,\;b=b,\;c=45$$, so we have

$$r_{1}+r_{2}=z+\bar z=2\alpha=-b\;\; \Longrightarrow\;\; \alpha=-\dfrac{b}{2}. \quad -(1)$$

and

$$r_{1}r_{2}=z\bar z=\alpha^{2}+\beta^{2}=45. \quad -(2)$$

The roots also satisfy the given geometric condition $$|z+1|=2\sqrt{10}.$$ For any complex number $$w=x+iy$$ we have the modulus formula $$|w|=\sqrt{x^{2}+y^{2}}.$$ Hence

$$|z+1|^{2}=((\alpha+1)+i\beta)((\alpha+1)-i\beta)=(\alpha+1)^{2}+\beta^{2}= (2\sqrt{10})^{2}=40. \quad -(3)$$

Now we possess two equations containing $$\alpha^{2}+\beta^{2}$$ and $$(\alpha+1)^{2}+\beta^{2}$$. Subtracting (2) from (3) eliminates $$\beta^{2}$$:

$$(\alpha+1)^{2}+\beta^{2}-(\alpha^{2}+\beta^{2})=40-45.$$ $$\bigl(\alpha^{2}+2\alpha+1\bigr)-\alpha^{2}= -5.$$ $$2\alpha+1=-5.$$ $$2\alpha=-6.$$ $$\alpha=-3. \quad -(4)$$

Using (1) we connect $$\alpha$$ to $$b$$:

$$-\dfrac{b}{2}=-3\;\;\Longrightarrow\;\; b=6. \quad -(5)$$

With the value of $$b$$ obtained, we examine the option statements. Computing $$b^{2}-b$$ gives

$$b^{2}-b=6^{2}-6=36-6=30. \quad -(6)$$

The option matching equation (6) is A. $$b^{2}-b=30.$$

All other options are easily checked to be inconsistent: $$b^{2}+b=42\neq72,\, b^{2}-b\neq42,\, b^{2}+b\neq12.$$

Hence, the correct answer is Option A.

We are asked to evaluate $$(2+\alpha)^4$$ where $$\alpha=\dfrac{-1+i\sqrt3}{2}$$ and then write the result in the form $$a+b\alpha$$ with real numbers $$a$$ and $$b$$. Afterwards we will calculate $$a+b$$.

First recall that $$\alpha$$ is a non-real cube root of unity. Hence it satisfies the standard relations

$$\alpha^3 = 1$$   and   $$1+\alpha+\alpha^2 = 0.$$

From $$1+\alpha+\alpha^2=0$$ we obtain the very useful identity

$$\alpha^2 = -1-\alpha.$$

Now we expand $$(2+\alpha)^4$$ with the Binomial Theorem. The theorem states that for any numbers $$x$$ and $$y$$,

$$ (x+y)^4=\;^4C_0x^4+\;^4C_1x^3y+\;^4C_2x^2y^2+\;^4C_3xy^3+\;^4C_4y^4,$$

where the binomial coefficients for the fourth power are $$^4C_0=1,\;^4C_1=4,\;^4C_2=6,\;^4C_3=4,\;^4C_4=1.$$

Putting $$x=2$$ and $$y=\alpha$$, we get

$$ (2+\alpha)^4 = 2^4\;+\;4\cdot2^3\alpha\;+\;6\cdot2^2\alpha^2\;+\;4\cdot2\alpha^3\;+\;\alpha^4. $$

Compute each numerical power of 2 first:

$$2^4=16,\qquad2^3=8,\qquad2^2=4.$$

Substituting these values,

$$ (2+\alpha)^4 = 16 \;+\; 4\cdot8\,\alpha \;+\; 6\cdot4\,\alpha^2 \;+\; 4\cdot2\,\alpha^3 \;+\; \alpha^4, $$

which simplifies to

$$ (2+\alpha)^4 = 16 \;+\; 32\alpha \;+\; 24\alpha^2 \;+\; 8\alpha^3 \;+\; \alpha^4. $$

We now reduce all higher powers of $$\alpha$$ using the identities already stated.

First, $$\alpha^3=1,$$ so

$$8\alpha^3 = 8\cdot1 = 8.$$

Next, $$\alpha^2=-1-\alpha,$$ so

$$24\alpha^2 = 24(-1-\alpha) = -24 - 24\alpha.$$

Finally, $$\alpha^4 = \alpha\cdot\alpha^3 = \alpha\cdot1 = \alpha.$$

Substituting each of these back into the expansion:

$$ (2+\alpha)^4 = 16 \;+\; 32\alpha \;+\;(-24 - 24\alpha) \;+\; 8 \;+\; \alpha. $$

Now collect the constant (real) terms and the coefficients of $$\alpha$$ separately.

Constant terms: $$16 + (-24) + 8 = 0.$$

$$\alpha$$ terms: $$32\alpha + (-24\alpha) + \alpha = (32-24+1)\alpha = 9\alpha.$$

Hence we have obtained the very compact result

$$ (2+\alpha)^4 = 0 + 9\alpha = 9\alpha. $$

In other words, $$a=0$$ and $$b=9.$$ Therefore,

$$a+b = 0+9 = 9.$$

Hence, the correct answer is Option A.

We have the complex expression $$z=\dfrac{3+i\sin\theta}{4-i\cos\theta},\qquad\theta\in[0,\,2\pi].$$

Because $$z$$ is given to be a real number, its imaginary part must vanish. A convenient way to isolate the imaginary part is to multiply the numerator and the denominator by the conjugate of the denominator, making the new denominator purely real.

The conjugate of $$\,4-i\cos\theta\,$$ is $$\,4+i\cos\theta.$$ Multiplying, we get

$$$z=\dfrac{3+i\sin\theta}{4-i\cos\theta}\cdot\dfrac{4+i\cos\theta}{4+i\cos\theta} =\dfrac{(3+i\sin\theta)(4+i\cos\theta)} {(4-i\cos\theta)(4+i\cos\theta)}.$$$

In the denominator we use the identity $$(a-b)(a+b)=a^{2}-b^{2}$$ with $$a=4$$ and $$b=i\cos\theta$$:

$$$(4-i\cos\theta)(4+i\cos\theta)=4^{2}-(i\cos\theta)^{2}=16-(-\cos^{2}\theta)=16+\cos^{2}\theta.$$$

Thus the denominator is the real number $$16+\cos^{2}\theta.$$ Now we concentrate on the numerator

$$(3+i\sin\theta)(4+i\cos\theta).$$

We expand it term by term:

$$$\begin{aligned} (3+i\sin\theta)(4+i\cos\theta) &=3\cdot4+3\cdot(i\cos\theta)+i\sin\theta\cdot4+i\sin\theta\cdot(i\cos\theta)\\ &=12+3i\cos\theta+4i\sin\theta+i^{2}\sin\theta\cos\theta\\ &=12+3i\cos\theta+4i\sin\theta-\sin\theta\cos\theta\\ &=(12-\sin\theta\cos\theta)+i\bigl(3\cos\theta+4\sin\theta\bigr). \end{aligned}$$$

The denominator is real, so for $$z$$ itself to be real, the imaginary part of the numerator must be zero:

$$3\cos\theta+4\sin\theta=0.$$

Dividing both sides by $$\cos\theta$$ (which is permissible wherever $$\cos\theta\neq0$$), we obtain

$$3+4\tan\theta=0\quad\Longrightarrow\quad\tan\theta=-\frac{3}{4}.$$

The acute angle whose tangent is $$\dfrac{3}{4}$$ is $$\alpha=\tan^{-1}\!\left(\frac{3}{4}\right).$$ Since the tangent is negative, $$\theta$$ must lie in the second or fourth quadrants. Hence

$$$\theta=\pi-\alpha\quad\text{or}\quad\theta=2\pi-\alpha,\qquad \text{where } \alpha=\tan^{-1}\!\left(\frac{3}{4}\right).$$$

Now we must find an argument of the complex number $$w=\sin\theta+i\cos\theta.$$

First observe the trigonometric identities

$$$\cos\!\left(\frac{\pi}{2}-\theta\right)=\sin\theta,\qquad \sin\!\left(\frac{\pi}{2}-\theta\right)=\cos\theta.$$$

Therefore

$$$w=\sin\theta+i\cos\theta =\cos\!\left(\frac{\pi}{2}-\theta\right)+i\sin\!\left(\frac{\pi}{2}-\theta\right) =e^{\,i\left(\frac{\pi}{2}-\theta\right)}.$$$

Thus one argument of $$w$$ is

$$\arg w=\frac{\pi}{2}-\theta.$$ Substituting each permissible $$\theta$$:

For $$\theta=\pi-\alpha$$ we get

$$$\arg w=\frac{\pi}{2}-(\pi-\alpha)=\frac{\pi}{2}-\pi+\alpha =-\frac{\pi}{2}+\alpha.$$$

For $$\theta=2\pi-\alpha$$ we get

$$$\arg w=\frac{\pi}{2}-(2\pi-\alpha)=\frac{\pi}{2}-2\pi+\alpha =-\frac{3\pi}{2}+\alpha.$$$

The two expressions differ by exactly $$-\,\pi$$, so they represent the same set of arguments (they vary by an integral multiple of $$\,\pi$$). To compare with the options, it is convenient to add $$\,\pi$$ to the first value:

$$-\frac{\pi}{2}+\alpha+\pi=\frac{\pi}{2}+\alpha =\pi-\bigl(\tfrac{\pi}{2}-\alpha\bigr).$$

But $$\tfrac{\pi}{2}-\alpha=\tan^{-1}\!\left(\frac{4}{3}\right)$$ because for complementary acute angles we have $$\tan\beta=\frac{1}{\tan\alpha}$$. Hence

$$\arg w=\pi-\tan^{-1}\!\left(\frac{4}{3}\right).$$

This value appears exactly as Option A.

Hence, the correct answer is Option A.

We start by writing the complex number in the usual Cartesian form $$z = x + iy$$, where $$x$$ and $$y$$ are real numbers. The given condition is

$$\text{Re}\!\left(\dfrac{z-1}{2z+i}\right)=1.$$

First we express the numerator and denominator separately in terms of $$x$$ and $$y$$.

Numerator: $$z-1=(x+iy)-1=(x-1)+iy.$$

Denominator: $$2z+i=2(x+iy)+i=2x+2iy+i=2x+i(2y+1).$$

It is convenient to denote

$$a=2x,\qquad b=2y+1,$$

so that the denominator becomes $$a+ib$$. To extract the real part of a quotient we multiply the numerator and denominator by the complex conjugate of the denominator. Using the identity

$$\dfrac{p+iq}{r+is}=\dfrac{(p+iq)(r-is)}{(r+is)(r-is)}=\dfrac{(pr+qs)+i(qr-ps)}{r^{2}+s^{2}},$$

we have

$$\dfrac{(x-1)+iy}{a+ib}\;=\;\dfrac{\bigl[(x-1)+iy\bigr](a-ib)}{a^{2}+b^{2}}.$$

Carrying out the multiplication in the numerator:

$$\bigl[(x-1)+iy\bigr](a-ib)= (x-1)a -i b(x-1) + i y a - i^{2}yb.$$

Remembering that $$i^{2}=-1$$, the term $$-i^{2}yb$$ becomes $$+yb$$. So we collect real and imaginary parts:

Real part: $$(x-1)a + yb,$$

Imaginary part: $$\Bigl[-b(x-1)+ya\Bigr]i.$$

Hence

$$\dfrac{z-1}{2z+i}= \dfrac{(x-1)a+yb}{a^{2}+b^{2}}\;+\;i\,\dfrac{-b(x-1)+ya}{a^{2}+b^{2}}.$$

The real part of this fraction is therefore

$$\text{Re}\!\left(\dfrac{z-1}{2z+i}\right)=\dfrac{(x-1)a+yb}{a^{2}+b^{2}}.$$

The given condition tells us that this real part equals $$1$$, so

$$\dfrac{(x-1)a+yb}{a^{2}+b^{2}} = 1.$$

Cross-multiplying gives

$$ (x-1)a + yb = a^{2} + b^{2}. $$

Now we substitute back $$a=2x$$ and $$b=2y+1$$.

Left side:

$$ (x-1)(2x) + y(2y+1) = 2x(x-1) + y(2y+1). $$

Right side:

$$ a^{2}+b^{2} = (2x)^{2} + (2y+1)^{2} = 4x^{2} + (2y+1)^{2}. $$

So the equation becomes

$$2x(x-1) + y(2y+1) = 4x^{2} + (2y+1)^{2}.$$

We now expand every term.

Left side:

$$2x(x-1)=2x^{2}-2x,\qquad y(2y+1)=2y^{2}+y,$$

so

$$\text{Left side}=2x^{2}-2x+2y^{2}+y.$$

Right side:

$$4x^{2} + (2y+1)^{2}=4x^{2} + 4y^{2}+4y+1.$$

Setting left equal to right:

$$2x^{2}-2x+2y^{2}+y = 4x^{2}+4y^{2}+4y+1.$$

We now bring every term to one side to obtain zero on the other:

$$0 = 4x^{2}+4y^{2}+4y+1 - (2x^{2}-2x+2y^{2}+y).$$

Simplifying term by term:

$$0 = (4x^{2}-2x^{2}) + (4y^{2}-2y^{2}) + (4y-y) + (1-(-2x)),$$

$$0 = 2x^{2} + 2y^{2} + 3y + 2x + 1.$$

Dividing by $$2$$ to make the coefficients smaller,

$$x^{2} + y^{2} + x + \dfrac{3}{2}y + \dfrac{1}{2}=0.$$

This is recognisably the general equation of a circle. To see its centre and radius we complete the square in both $$x$$ and $$y$$.

For the $$x$$-terms:

$$x^{2}+x = x^{2}+x+\dfrac14 -\dfrac14 = \left(x+\dfrac12\right)^{2}-\dfrac14.$$

For the $$y$$-terms:

$$y^{2}+\dfrac{3}{2}y = y^{2}+\dfrac{3}{2}y+\dfrac{9}{16}-\dfrac{9}{16}= \left(y+\dfrac34\right)^{2}-\dfrac{9}{16}.$$

Substituting these back we have

$$\left(x+\dfrac12\right)^{2}-\dfrac14 \;+\; \left(y+\dfrac34\right)^{2}-\dfrac{9}{16} \;+\; \dfrac12 = 0.$$

The constants combine as

$$-\dfrac14 -\dfrac{9}{16} + \dfrac12 = -\dfrac{4}{16}-\dfrac{9}{16}+\dfrac{8}{16}=-\dfrac{5}{16}.$$

Transferring this constant to the right-hand side gives the standard form

$$\left(x+\dfrac12\right)^{2} + \left(y+\dfrac34\right)^{2} = \dfrac{5}{16}.$$

Thus the locus is a circle with centre $$\left(-\dfrac12,\;-\dfrac34\right)$$ and radius $$\sqrt{\dfrac{5}{16}}=\dfrac{\sqrt5}{4}.$$ Its diameter is twice the radius, namely

$$2\times\dfrac{\sqrt5}{4}=\dfrac{\sqrt5}{2}.$$

This matches the description of Option D: “circle whose diameter is $$\dfrac{\sqrt{5}}{2}$$.”

Hence, the correct answer is Option D.

Let us write the unknown complex number in the usual form $$z=x+iy$$, where $$x=\text{Re}(z)$$ and $$y=\text{Im}(z)$$. Its conjugate is therefore $$\overline z=x-iy$$ and, very importantly, $$\text{Re}(z)=\text{Re}(\overline z)=x$$.

Now we list the four given vertices in coordinate form (treating a complex number $$a+ib$$ as the point $$(a,b)$$ in the Argand plane):

$$\begin{aligned} A&:z &&\;\;\;\;=(x,y),\\ B&:\overline z &&\;\;\;\;=(x,-y),\\ C&:\overline z-2\,\text{Re}(\overline z)&&\;\;\;\;=\bigl(x-2x,\,-y\bigr)=(-x,-y),\\ D&:z-2\,\text{Re}(z) &&\;\;\;\;=\bigl(x-2x,\,y\bigr)=(-x,y). \end{aligned}$$

So the four points are $$A(x,y),\;B(x,-y),\;C(-x,-y),\;D(-x,y).$$ One can notice immediately that these points are the four corners of a rectangle centred at the origin; the problem tells us it is actually a square.

We invoke the distance formula between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$:

$$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$

Taking consecutive vertices $$A(x,y)$$ and $$B(x,-y)$$, we have

$$AB=\sqrt{(x-x)^2+(-y-y)^2}=\sqrt{0^2+( -2y)^2}=2|y|.$$

Similarly, for the next side $$BC$$ between $$B(x,-y)$$ and $$C(-x,-y)$$ we get

$$BC=\sqrt{(-x-x)^2+(-y+ y)^2}=\sqrt{(-2x)^2+0^2}=2|x|.$$

Because all four sides of a square are equal, we must have

$$2|y|=2|x|\;\;\Longrightarrow\;\;|x|=|y|.$$

The problem states that each side of the square measures 4 units, so

$$2|y|=4\;\;\Longrightarrow\;\;|y|=2.$$

Using $$|x|=|y|$$ immediately gives

$$|x|=2.$$

Finally, the modulus of $$z$$ is obtained from the definition $$|z|=\sqrt{x^2+y^2}:$$

$$|z|=\sqrt{x^2+y^2}=\sqrt{(2)^2+(2)^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt2.$$

Hence, the correct answer is Option 3.