"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 __________."

Question

Accepted Answer

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$z = 1 + i$

${z_1} = {{1 + i\overline z } \over {\overline z (1 - z) + {1 \over z}}}$

$= {{z(1 + i\overline z )} \over {|z{|^2}(1 - z) + 1}}$

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

${z_1} = 1 - i$

$\arg {z_1} = {\tan ^{ - 1}}\left( {{{ - 1} \over 1}} \right) = {{3\pi } \over 4}$

${{12} \over \pi }\arg ({z_1}) = {{3\pi } \over 4}\,.\,{{12} \over \pi } = 9$

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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 __________.