Algorithmic/procedural way to find branch points and branch cuts?

Question

I am reading a complex analysis book. I am trying to understand if there is an algorithmic/procedural way to identify branch points and branch cuts.

For example, I have the function $ln(1-z^{1/2})$.

1. I know that $ln(0)$ is undefined. I set $1-z^{1/2}=0 \rightarrow z=1$. Does this mean that 1 is a branch point? Is it the only branch point?

2. Does this mean that the branch cut is $(-\infty,1) \cup (1, \infty)$? If not, how do I find the branch cut?

Thank you very much.

Answer 1

We consider two functions $w=1-z^{1/2}$ and $\log w$.
We define normally $$z^{1/2}=\sqrt{r}e^{i\theta /2} \quad(z=re^{i\theta },\, |\theta |<\pi), $$ so $z^{1/2}$ is defined for $z\in \mathbb{C}\setminus (-\infty,0\,]$ and $(-\infty,0\,]$ is the branch cut of $z^{1/2}$.
Then $z^{1/2}$ satisfies $\operatorname{Re }z^{1/2}>0$ since $|\arg (z^{1/2})|<\frac{\pi}{2}.$ From this we see $$ \operatorname{Re} w=\operatorname{Re }\left(1-z^{1/2}\right)<1.$$ On the other hand $\log w$ is defined normally as follows (the principal branch):$$ \log w=\log r+i\theta\quad (w=re^{i\theta }, |\theta |<\pi).$$ In other words, $\log w$ has a branch cut $(-\infty,0\,]$. We note that $-\infty<w=1-z^{1/2}\le 0$ corresponds to $1\le z^{1/2}$, which corresponds to $1\le z.$

Thus $\log (1-z^{1/2})$ has a branch cut $(-\infty, 0\,]\cup [\,1,\infty)$.