If $\left|z\right|=1, z\ne 1,$ then value of $arg\left(\frac{1}{1-z}\right)$ cannot exceed

As $\left|z\right|=1, z\ne 1, z=\cos \theta +i\sin {\theta }_{,}-\pi <\theta \le \pi , \theta \ne 0.$ Now 

                                                $\omega =\frac{1}{1-z}=\frac{1}{1-\cos \theta -i\sin \theta }$
                                         $\begin{array}{l}=\frac{(1-\cos \theta )+i\sin \theta }{(1-\cos \theta {)}^2+{\sin }^2\theta }=\frac{(1-\cos \theta )+i\sin \theta }{2(1-\cos \theta )}\\ =\frac{1}{2}+\frac{i}{2}\cot \left(\frac{\theta }{2}\right)=\frac{1}{2}+\frac{i}{2}\tan \left(\frac{\pi }{2}-\frac{\theta }{2}\right)\end{array}$

This shows that $w$ lies on the line $x =1/2$ and $-\pi /2<\arg \left(\omega \right)<\pi /2,\mathrm{Arg}\left(\omega \right)\ne 0,$ The maximum value of $Arg\left(w\right)$ is never attained.