For $z \in \mathbb{C}$, |z-1| = 1 proof that arg(z-1) = 2arg(z) = $\frac{2}{3}$arg($z^2-z$)

Question

I know that the first part of my question has been asked a year ago, but the solutions there don't help me, since I'm not familiar with the theorems they use there. So I'd be glad if someone had a basic approach to solve this.

The approach which lab bhattacharjee suggested is what I was looking for. I just need some help to develop it further, please.

Answer 1

Hint

WLOG arg$(z-1)=2t\implies z-1=\cos2t+i\sin2t$

$z=2\cos t(\cos t+i\sin t),$arg$(z)=?$

Now $z(z-1)=2\cos t(\cos t+i\sin t)(\cos2t+i\sin2t)=2\cos t(\cos3t+i\sin3t)$

Answer 2

This is a simple consequence of the inscribed angle theorem:

If $|z-1|=1$, the image $M$ of $z$ in the Argand-Cauchy plane is a point of the circle with center $I$ (the image of $1$) and its $\arg(z-1)$ is the polar angle of the straight line $(IM)$, i.e. a central angle on this circle, and $\arg z$ is a corresponding inscribed angle.