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)
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