<s>[INST] The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$ 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 $\frac{\pi}{2}$
 [/INST] $$\left| {\arg\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$$ $$= \left|\arg \left(\operatorname{cis}\left(\frac{\pi - \theta}{2}\right)\right) - \arg\left(2\sin \frac{\theta }{2}\right)\right|$$
$$=\left|\arg \left(\operatorname{cis}\left(\frac{\pi - \theta}{2}\right)\right)\right| = \left|\frac{\pi-\theta}{2} \right| \leq \frac{\pi}{2}$$
Here, $\operatorname{cis}(x) = \cos x + i\sin x$.
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