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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.

MathIsHard
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1 Answers1

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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 $(-\infty, 0\,]\cup [\,1,\infty)$.

ts375_zk26
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  • If I plug in 1/2 I get a result on the negative axis. Wouldn't that mean the cut isn't quite right? – MathIsHard Mar 31 '17 at 00:40
  • If we plug in $1/2$ we get a result $\log(1-(1/2)^{1/2})=\log (1-\sqrt{2}/2)\approx -1.22.$ The value is negative, since $0<1-\sqrt{2}/2<1$. It is not a strange result. Important is that $\log(1-z^{1/2})$ is single-valued on $\mathbb{C}\setminus (-\infty, 0,]\cup [1,\infty)$. – ts375_zk26 Mar 31 '17 at 02:18
  • Thank you. I think also that it is equal to 0.53 for -pi/2 <arg< pi/2 so it works out for that arg restriction. I appreciate the help. – MathIsHard Mar 31 '17 at 02:21
  • Oh. I see what you mean. As long as it is single valued it doesn't matter where the outcome is located. I was getting caught up in the wrong thing I think – MathIsHard Mar 31 '17 at 02:48
  • Why must 2 be cut out of the branch? – MathIsHard Mar 31 '17 at 02:53
  • If we define $$\log w=\log r+i\theta\quad (w=re^{i\theta }, |\theta |<\pi),$$ then $\log w$ is not defined for real negative values of $w$. So the value of $\log (1-2^{1/2})$ doesn't exist. $2$ must be cut out. If we define $$\log w=\log r+i\theta\quad \left(w=re^{i\theta }, -\frac{\pi}{2}<\theta <\frac{3\pi}{2}\right),$$ then $\log(1-2^{1/2})=\log(\sqrt{2}-1)+i\pi.$ In this case $2$ is not cut out of the branch. – ts375_zk26 Mar 31 '17 at 03:19