Argand Plane

Given,

$$\left|\frac{z-2}{z+2}\right|=1$$

Hence,

$$|z-2|=|z+2|$$

Let

$$z=x+iy$$

Then,

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

Squaring both sides,

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

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

$$x=0$$

Thus,

$$\operatorname{Re}(z)=0$$

Now,

$$|z|^2-4\operatorname {Re}(z)= |z|^2$$

Since

$$z=iy,$$

$$|z|^2=y^2$$

which is not fixed.

Now observe geometrically:

$$|z-2|=|z+2|$$

represents the perpendicular bisector of

$$2$$

and

$$-2,$$

that is,

$$x=0$$

Hence,

$$|z|^2-4\operatorname{Re}(z)=x^2+y^2-4x$$

$$=y^2$$

Taking the least possible value,

$$y=0$$

Thus minimum value is

$$0$$

Therefore, the required answer is

$$\boxed0$$

Let the denominator be $$D = \bar{z}(1 - z) + \frac{1}{z}$$.

Given $$z = 1 + i$$, then $$\bar{z} = 1 - i$$.

• Part 1: $$\bar{z}(1 - z) = (1 - i)(1 - (1 + i)) = (1 - i)(-i) = -i + i^2 = -1 - i$$
• Part 2: $$\frac{1}{z} = \frac{1}{1 + i}$$. Multiply by conjugate: $$\frac{1-i}{(1+i)(1-i)} = \frac{1-i}{2} = \frac{1}{2} - \frac{i}{2}$$

Combine them:

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

Let the numerator be $$N = 1 + i\bar{z}$$.

$$N = 1 + i(1 - i) = 1 + i - i^2 = 1 + i + 1 = 2 + i$$

$$z_1 = \frac{N}{D} = \frac{2 + i}{\frac{-1 - 3i}{2}} = \frac{2(2 + i)}{-(1 + 3i)}$$

Multiply the numerator and denominator by the conjugate of the denominator $$(1 - 3i)$$:

$$z_1 = \frac{-2(2 + i)(1 - 3i)}{(1 + 3i)(1 - 3i)} = \frac{-2(2 - 6i + i - 3i^2)}{1 + 9}$$ $$z_1 = \frac{-2(2 - 5i + 3)}{10} = \frac{-2(5 - 5i)}{10} = \frac{-10(1 - i)}{10}$$ $$z_1 = -1 + i$$

The complex number $$z_1 = -1 + i$$ lies in the Second Quadrant (Real part is negative, Imaginary part is positive).

$$\arg(z_1) = \pi - \tan^{-1}\left(\left|\frac{1}{-1}\right|\right)$$ $$\arg(z_1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$

Substitute $$\arg(z_1)$$ into the required expression:

$$\text{Value} = \frac{12}{\pi} \times \left(\frac{3\pi}{4}\right)$$ $$\text{Value} = 3 \times 3 = 9$$

Correct Option: C (9)

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

From $$|z - 1| = 1$$: $$|x - 1 + iy| = 1$$

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

This is a circle centered at $$(1, 0)$$ with radius 1.

From $$|z - i| = 1$$: $$|x + i(y-1)| = 1$$

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

This is a circle centered at $$(0, 1)$$ with radius 1.

Finding the intersection of the two circles:

From equation 1: $$x^2 - 2x + 1 + y^2 = 1 \Rightarrow x^2 + y^2 = 2x$$

From equation 2: $$x^2 + y^2 - 2y + 1 = 1 \Rightarrow x^2 + y^2 = 2y$$

Therefore: $$2x = 2y \Rightarrow x = y$$

Substituting into $$x^2 + y^2 = 2x$$:
$$x^2 + x^2 = 2x$$
$$2x^2 = 2x$$
$$x^2 = x$$
$$x(x-1) = 0$$

So $$x = 0$$ or $$x = 1$$

If $$x = 0$$: $$y = 0$$, so $$z = 0$$

Check: $$|0-1| = 1$$  and $$|0-i| = 1$$ 

If $$x = 1$$: $$y = 1$$, so $$z = 1 + i$$

Check: $$|1+i-1| = |i| = 1$$  and $$|1+i-i| = |1| = 1$$ 

For $$z = 0$$: $$|z| = 0$$
For $$z = 1 + i$$: $$|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$$

since  $$\sqrt{2}$$ is only there among 2 solutions, $$\sqrt{2}$$ is the answer