Find the imaginary part of $\ln(1-re^{ix})$

Question

I'm trying to prove that imaginary part of $\ln(1-re^{ix})$ is $$\Im \left(\ln(1-re^{ix})\right)=\arctan\left(\frac{r\sin(x)}{1-r\cos(x)}\right).$$

I saw this assertation in the book "Trigonometric Series"

Answer 1

Let $z = |z|e^{i\alpha} = 1 - re^{ix} = (1 - r\cos x) -ir\sin x$. Note that $$\log z = \log\left(|z|e^{i\alpha}\right) = \log |z| + i\alpha$$

whose imaginary part is $\alpha$. So we need only evaluate $\alpha$. But $\alpha = \arg(1-re^{ix})$ and we have $1-re^{ix}= (1-r\cos x) - ir\sin x$, from which we can conclude $\alpha = \arctan \frac{r\sin x}{1-r\cos x}$ under suitable restrictions.