JEE MAIN - Mathematics (2023 - 30th January Morning Shift - No. 13)
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
9
Explanation
$$z = 1 + i$$
$${z_1} = {{1 + i\overline z } \over {\overline z (1 - z) + {1 \over z}}}$$
$$ = {{z(1 + i\overline z )} \over {|z{|^2}(1 - z) + 1}}$$
$$ = {{(1 + i)(1 + i(1 - i))} \over {2(1 - 1 - i) + 1}}$$
$${z_1} = 1 - i$$
$$\arg {z_1} = {\tan ^{ - 1}}\left( {{{ - 1} \over 1}} \right) = {{3\pi } \over 4}$$
$${{12} \over \pi }\arg ({z_1}) = {{3\pi } \over 4}\,.\,{{12} \over \pi } = 9$$
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