Let z = 1 + i and z _ 1 = 1 + i ¯ z ¯ z ( 1 - z ) + (1)/(z) · Then...

Given:

$z = 1 + i$

$\overset{¯}{z} = 1 - i$

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

So,

${z}_{1} = \dfrac{1 + i\overset{¯}{z}}{\overset{¯}{z} ( 1 - z ) + \dfrac{1}{z}}$

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

$= \dfrac{1 + i - {i}^{2}}{- i + {i}^{2} + \dfrac{1}{2} - \dfrac{i}{2}}$

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

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

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

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

Arg \left({z}_{1}\right) = \dfrac{3\pi}{4}$</p><p>$∴ \dfrac{12}{\pi}arg\left({z}_{1}\right) = \dfrac{12}{\pi} \times \dfrac{3\pi}{4} = 9$