Let z = 1 + i and z 1 = 1 + i z ¯ z ¯ ( 1 - z ) + 1 z · Then 12 π arg z 1 is equal to

Given:

$z=1+i$

$\overline{z}=1-i$

And, $\frac{1}{z}=\frac{1}{1+i}\times \frac{1-i}{1-i}=\frac{1-i}{2}$

So,

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

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

$=\frac{1+i-i^2}{-i+i^2+\frac{1}{2}-{\displaystyle \frac{i}{2}}}$

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

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

$=\frac{12i-4+6i^2-2i}{10}$

$=\frac{10i-10}{10}=i-1$

$Arg\left(z_1\right)=\frac{3\pi }{4}$

$\therefore \frac{12}{\pi }\arg \left({\mathrm{z}}_1\right)=\frac{12}{\pi }\times \frac{3\pi }{4}=9$