4-1.Complex numbers
hard

Let $z =1+ i$ and $z _1=\frac{1+ i \overline{ z }}{\overline{ z }(1- z )+\frac{1}{ z }}$. Then $\frac{12}{\pi}$ $\arg \left(z_1\right)$ is equal to $..........$.

A

$18$

B

$27$

C

$36$

D

$9$

(JEE MAIN-2023)

Solution

$z=1+i$

$z_1=\frac{1+i \bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$

$z_1=\frac{1+i(1-i)}{(1-i)(1-1-i)+\frac{1}{1+i}}$

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

$=\frac{2+i}{-3 i-1}=\frac{4+2 i}{-3 i-1}$

$=\frac{-(4+2 i)(3 i-1)}{(3 i)^2-(1)^2}$

$\therefore \frac{12}{\pi} \arg \left(z_1\right)=\frac{12}{\pi} \times \frac{3 \pi}{4}=9$

Standard 11
Mathematics

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