Complex Number

Since

$$\omega=\cos\frac{2\pi}{9}+i\sin\frac{2\pi}{9},$$

we have

$$\omega^9=1$$

and

$$\omega\ne1$$

Using the sum of GP,

$$1+\omega+\omega^2+\cdots+\omega^8=0$$

Hence,

$$\omega+\omega^2+\omega^3+\omega^4 =(1+\omega^5+\omega^6+\omega^7+\omega^8)$$

Now take real parts.

Using

$$\operatorname{Re}(\omega^k)=\cos\frac{2k\pi}{9},$$

we get

$$\operatorname{Re}(\omega+\omega^2+\omega^3+\omega^4)$$

$$=-(1+\cos\frac{10\pi}{9}+\cos\frac{12\pi}{9}+\cos\frac{14\pi}{9}+\cos\frac{16\pi}{9})$$

Using symmetry of cosine roots of unity,

$$\cos\frac{2\pi}{9}+\cos\frac{4\pi}{9}+\cos\frac{6\pi}{9}+\cos\frac{8\pi}{9}=-\frac12$$

Hence,

$$\operatorname{Re}(\omega+\omega^2+\omega^3+\omega^4)= -\left(1-\frac12\right)$$

$$=-\frac12$$

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)