Conjugate of a Complex Number

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)

To find the required value, we use the properties of roots of unity, complex rotation, and conjugates step-by-step:

1. Determine the vertices of the equilateral triangle:
The vertices $$z_1, z_2, z_3$$ are inscribed in the circle $$|z| = 2$$ and form an equilateral triangle. Since $$z_1 = 2$$ lies on the positive real axis, the other two vertices are obtained by rotating $$z_1$$ counter-clockwise about the origin by angles of $$\frac{2\pi}{3}$$ and $$\frac{4\pi}{3}$$ respectively.
Using the complex cube root of unity $$\omega = e^{i\frac{2\pi}{3}}$$, we can write:
$$z_1 = 2$$
$$z_2 = 2\omega$$
$$z_3 = 2\omega^2$$

2. Calculate the conjugate of $$z_3$$:
We find the conjugate of $$z_3 = 2\omega^2$$. Since $$\bar{\omega} = \omega^2$$ and $$\overline{\omega^2} = \omega$$, the conjugate simplifies to:
$$\bar{z_3} = \overline{2\omega^2} = 2\omega$$

3. Evaluate the transformed complex number $$z_4$$:
Substitute the values of $$z_1$$, $$z_2$$, and $$\bar{z_3}$$ into the given expression:
$$z_4 = \frac{z_2 \cdot \bar{z_3}}{z_1}$$
$$z_4 = \frac{(2\omega) \cdot (2\omega)}{2}$$
$$z_4 = \frac{4\omega^2}{2} = 2\omega^2$$

Notice that $$z_4$$ is exactly equal to the vertex $$z_3$$.

4. Compute the final absolute squared value:
We need to find the value of $$|z_4 - z_1|^2$$, which is equivalent to finding the squared distance between the vertices $$z_3$$ and $$z_1$$ of our equilateral triangle.
$$|z_4 - z_1|^2 = |2\omega^2 - 2|^2 = 4|\omega^2 - 1|^2$$

We substitute the algebraic value of $$\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$$ into the magnitude expression:
$$|\omega^2 - 1| = \left| -\frac{1}{2} - i\frac{\sqrt{3}}{2} - 1 \right| = \left| -\frac{3}{2} - i\frac{\sqrt{3}}{2} \right|$$
$$|\omega^2 - 1|^2 = \left(-\frac{3}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{9}{4} + \frac{3}{4} = \frac{12}{4} = 3$$

Now, multiply this by the scalar factor of 4:
$$|z_4 - z_1|^2 = 4 \times 3 = 12$$

Geometrically, the side length $$s$$ of an equilateral triangle inscribed in a circle of radius $$R = 2$$ is given by $$s = R\sqrt{3} = 2\sqrt{3}$$. The square of the side length (the distance between $$z_3$$ and $$z_1$$) is $$s^2 = (2\sqrt{3})^2 = 12$$.

First simplify

$$\frac{1+i}{1-i}$$

Multiply numerator and denominator by the conjugate

$$1+i$$

$$\frac{1+i}{1-i}\cdot\frac{1+i}{1+i}$$

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

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

$$=\frac{2i}{2}$$

$$=i$$

Therefore,

$$Z=i^3$$

Using

$$i^2=-1$$

$$i^3=i^2\cdot i$$

$$=(-1)i$$

$$=-i$$

Final Answer :

$$-i$$