Let z=1+i and z1=1+iz¯z¯(1−z)+1z. Then 12πarg(z1) is equal to____.

z1=1+i(1−i)(1−i)(1−1−i)+11+i=1+i−i2(1−i)(−i)+1−i2=2+i−3i−12=4+2i−3i−1z1=2(2+i)(1−3i)−(1+9)=−210((2+3)+i(1−6))z1==1−i12πarg(z1)=12π×3π4=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 $\frac{12}{π} \arg (z_{1})$ is equal to____.

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answer is 9.

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$z_{1} = \frac{1 + i (1 - i)}{(1 - i) (1 - 1 - i) + \frac{1}{1 + i}} = \frac{1 + i - i^{2}}{(1 - i) (- i) + \frac{1 - i}{2}}$

$= \frac{2 + i}{\frac{- 3 i - 1}{2}} = \frac{4 + 2 i}{- 3 i - 1}$

$z_{1} = \frac{2 (2 + i) (1 - 3 i)}{- (1 + 9)} = \frac{- 2}{10} ((2 + 3) + i (1 - 6))$

$z_{1} = = 1 - i$
$\begin{array}{l} \frac{12}{π} arg (z_{1}) = \frac{12}{π} \times \frac{3 π}{4} = 9 \end{array}$

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