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

The correct answer is 9.
z−1+i
z1​=zˉ(1−z)+z1​1+izˉ​
z1​=(1−i)(1−1−i)+1+i1​1+i(1i)​
=(1−i)(−i)+21−i​1+i−i2​
=−3i−12+i​=−3i−14+2i​
=(3i)2−(1)2−(4+2i)(3i−1)​
Arg(z1​)=43π​
∴π12​arg(z1​)=43π​