Find the maximum value of $|\arg (\frac{1}{1-z})|$ for $|z|=1$ $$\arg \left(\frac{1}{1-z}\right)$$
$$=\arg (1) - \arg (1-z)$$
$$=-\arg (1-z)$$
Placing the modulus gives
$$|\arg (1-z)|$$
Since it’s a circle, one point is $(1,0)$, then the point which is farthest away is $(-1,0)$, so the arg should be $\pi$. The correct answer is $\frac{\pi}{2}$. How is true?
I think I went wrong in using $\arg (1-z)$ when should be $\arg (z-1)$. I am not sure if that changes things, but that’s a possible flaw I noticed.

let $z=e^{ia}$
$1-z=2sin(a/2)e^{ia/2}$
thus
$arg(1-z)=a/2$
now by conventional domain $0\le a\le \pi$ and substituite...
