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Q.
If $|z|=1, z \neq 1$, then value of $\arg \left(\frac{1}{1-z}\right)$ cannot exceed
Complex Numbers and Quadratic Equations
Solution:
As $|z|=1, z \neq 1, z=\cos \theta+i \sin \theta_5-\pi<\theta$, $\leq \pi, \theta \neq 0$. Now
$\omega=\frac{1}{1-z}=\frac{1}{1-\cos \theta-i \sin \theta}$
$=\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)} $
$=\frac{1}{2}+\frac{i}{2} \cot \left(\frac{\theta}{2}\right)=\frac{1}{2}+\frac{i}{2} \tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)$
This shows that $\omega$ lies on the line $x=1 / 2$ and $-\pi / 2$ $<\arg (\omega)<\pi / 2, \operatorname{Arg}(\omega) \neq 0$,
The maximum value of $\operatorname{Arg}(\omega)$ is never attained.