JEE Main 2023 (Online) 30th January Morning ShiftQuestion 1
Let $z=1+i$ and $z_1=\frac{1+i\bar z}{\bar z(1-z)+\frac1z}$. Then $\frac{12}{\pi}\arg z_1$ is equal to ______.
Complex Numbers
Answer$9$
Solution
$1+i\bar z=2+i$, $\bar z(1-z)+\frac1z=-1-i+\frac{1-i}2=-\frac{1+3i}2$. $z_1=\frac{-2(2+i)}{1+3i}=-1+i$, $\arg z_1=\frac{3\pi}4$, so $\frac{12}\pi\arg z_1=9$.