Question

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

Answer 1

Answer: 9

$z-1+i$
$z_1=\frac{1+i\bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$
$z_1=\frac{1+i(1i)}{(1-i)(1-1-i)+\frac{1}{1+i}}$
$=\frac{1+i-i^2}{(1-i)(-i)+\frac{1-i}{2}}$
$=\frac{2+i}{-3i-1}=\frac{4+2i}{-3i-1}$
$=\frac{-(4+2i)(3i-1)}{(3i{)}^2-(1{)}^2}$
$\text{Arg}\left(z_1\right)=\frac{3\pi }{4}$
$\therefore \frac{12}{\pi }\arg \left(z_1\right)=\frac{12}{\pi }\times \frac{3\pi }{4}=9$