Let z = 1 + i and z_1 = frac{1 + i overline{z | vedclass

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(D) Given $z = 1 + i$,so $\overline{z} = 1 - i$ and $\frac{1}{z} = \frac{1}{1 + i} = \frac{1 - i}{2}$.Substitute these into the expression for $z_1$:$

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$9$

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(D) Given $z = 1 + i$,so $\overline{z} = 1 - i$ and $\frac{1}{z} = \frac{1}{1 + i} = \frac{1 - i}{2}$.Substitute these into the expression for $z_1$:$z_1 = \frac{1 + i(1 - i)}{(1 - i)(1 - (1 + i)) + \frac{1 - i}{2}}$$z_1 = \frac{1 + i - i^2}{(1 - i)(-i) + \frac{1 - i}{2}}$$z_1 = \frac{1 + i + 1}{-i + i^2 + \frac{1 - i}{2}} = \frac{2 + i}{-i - 1 + \frac{1 - i}{2}}$$z_1 = \frac{2 + i}{\frac{-2i - 2 + 1 - i}{2}} = \frac{2(2 + i)}{-1 - 3i} = \frac{2(2 + i)}{-(1 + 3i)}$Multiply numerator and denominator by $(1 - 3i)$:$z_1 = \frac{-2(2 + i)(1 - 3i)}{1^2 + 3^2} = \frac{-2(2 - 6i + i - 3i^2)}{10} = \frac{-2(2 - 5i + 3)}{10} = \frac{-2(5 - 5i)}{10} = -(1 - i) = -1 + i$.Now,find $\arg(z_1)$ for $z_1 = -1 + i$:Since $z_1$ is in the second quadrant,$\arg(z_1) = \pi - 	an^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Finally,calculate $\frac{12}{\pi} \arg(z_1)$:$\frac{12}{\pi} 	imes \frac{3\pi}{4} = 3 	imes 3 = 9$.

List-$I$ (Complex number)	List-$II$ (Polar form)
$(i) \sqrt{3}-i$	$(a) 2 \operatorname{cis} \frac{\pi}{6}$
$(ii) \sqrt{3}+i$	$(b) 2 \operatorname{cis} \frac{5 \pi}{6}$
$(iii) -\sqrt{3}+i$	$(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$
$(iv) -\sqrt{3}-i$	$(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$

Let $z = 1 + i$ and $z_1 = \frac{1 + i \overline{z}}{\overline{z}(1 - z) + \frac{1}{z}}$. Then $\frac{12}{\pi} \arg(z_1)$ is equal to $..........$.

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