Question

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

• A. $\pi /2$
• B. $\pi$
• C. $3\pi /2$
• D. $2\pi$

Answer 1

Answer: (A)

As $|z|=1,z\ne 1,z=\cos \theta +i\sin {\theta }_5-\pi <\theta$, $\le \pi ,\theta \ne 0$. 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 $-\pi /2$ $<\arg (\omega )<\pi /2,\mathrm{Arg}(\omega )\ne 0$,
The maximum value of $\mathrm{Arg}(\omega )$ is never attained.