The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$

Question

The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$=_____

My approach is as follow

Already this question is solved Maximum value of argument but I would like to slve by my approach which is illustrated below

$z = \cos \theta + i\sin \theta ;\theta \notin \left\{ 0 \right\}$ because $z\ne1$

$\left| {\operatorname{Arg}\left( {\frac{1}{{1 - \cos \theta - i\sin \theta }}} \right)} \right|$

$\left| {\operatorname{Arg}\left( {\frac{1}{{2{{\sin }^2}\frac{\theta }{2} - 2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)} \right|$

$\left| {\operatorname{Arg}\left( {\frac{1}{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} - i\cos \frac{\theta }{2}} \right)}}} \right)} \right| \Rightarrow \left| {\operatorname{Arg}\left( {\frac{{\left( {\sin \frac{\theta }{2} + i\cos \frac{\theta }{2}} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$

$\left| {\operatorname{Arg}\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$

How do we get the required answer which if $\frac{\pi}{2}$

Answer 1

$$\left| {\arg\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$$ $$= \left|\arg \left(\operatorname{cis}\left(\frac{\pi - \theta}{2}\right)\right) - \arg\left(2\sin \frac{\theta }{2}\right)\right|$$

$$=\left|\arg \left(\operatorname{cis}\left(\frac{\pi - \theta}{2}\right)\right)\right| = \left|\frac{\pi-\theta}{2} \right| \leq \frac{\pi}{2}$$

Here, $\operatorname{cis}(x) = \cos x + i\sin x$.

Answer 2

Hint:

Arg$\left(\dfrac12+i\cdot\dfrac{\cot\dfrac\theta2}2\right)=$Arg$\left(\cot\dfrac\theta2\right)=$Arg$\left(\tan\left(\dfrac{\pi-\theta}2\right)\right)$

Use this