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.
 A: There is no maximum for the argument:
Let $z=e^{i\theta}=\cos\theta+i\sin\theta,$ then $1-z=2\sin(\theta/2)(\sin(\theta/2)-i\cos(\theta/2))$ and hence $$\dfrac{1}{1-z}=\dfrac{1}{2\sin(\theta/2)}(\sin(\theta/2)+i\cos(\theta/2))=\dfrac{e^{\frac{i}2\left(\pi-\theta\right)}}{2\sin(\theta/2)}$$ for $\theta\neq 0.$ As you can see, argument change is $$\theta\mapsto \dfrac{\pi}2-\dfrac{\theta}2$$ for any $\theta\in(0,2\pi).$
If you want to see this more geometrically, observe that the image of the unit circle under the (Mobius) transformation $\dfrac{1}{1-z}$ is line $\Re(z)=\dfrac12$ on which there is no complex number with a maximum (or minimum) argument.
A: Denote the points $Z,O,I$ corresponding to the complex numbers $z,0,1,$ respectively.
$$\arg \left(\frac{1}{1-z}\right)=\arg \left(\frac{1-0}{1-z}\right)=\arg \left(\frac{0-1}{z-1}\right)$$ is the oriented angle between vectors $\vec{IZ}$ and $\vec{IO}.$
The angle gets larger when $Z$ approaches $I$ (on the unit circle) from "above" or, better said, clockwise. The value of the angle approaches $\pi/2,$ but this is never obtained.
Absolute value of the argument doesn't give more (see picture): approaching $I$ counterclockwise gives $$\left|\arg \left(\frac{0-1}{z-1}\right)\right|\to |-\pi/2|.$$

A: 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...
