If a function $f(z)$ has a branch point around $\alpha$ than the endpoints of a path around $\alpha$ do not map to the same point under $f(z)$. So we must test for inequality of $f(\alpha + re^{i2\pi}) \neq f(\alpha + re^{0})$. But this method only works for simple functions. For example given a fucntion

$$g(z) = \sqrt{1 + \sqrt{z}}$$

Zero is obviously a branch poin, but 1 is also a branch point for (2 out of 4) branches of this function where the inner square root is the branch for which $\sqrt{1} = -1$. I understand why this is, but I don't understand how to algebraically prove/show it?


Rewrite $w=\sqrt{1+\sqrt{z}}$ as $z=(w^2-1)^2$, and note where the derivative of the latter wrt $w$ is zero. You get $4w(w^2-1)=0$, so the branch points correspond to $w=0$ ($z=1$) and $w=\pm1$ ($z=0$).

  • $\begingroup$ Could you please explain why this works? $\endgroup$ – Arkya Chatterjee Aug 31 '16 at 10:30
  • $\begingroup$ @ArkyaChatterjee In this case, we have $z=f(w)$ for some analytic function $f$, and we are interested in branch points of the inverse. Note that near any point with $f'(w)≠0$, $f$ has a local analytic inverse, so there is no branch point. But if $f'(w_0)=0$, we can write $f(w)=a_0+(w-w_0)^ng(w)$ for some constant $a_0$ and analytic $g$ with $g(w_0)≠0$, and $n>1$. If $w$ traverses a small circle around $w_0$, so $g(w)$ is near constant, then $z=f(w)$ will go $n$ times around $f(w_0)$. And having $z$ go $n$ times around a point in order to make $w$ go once around, is the hallmark of a branch pt. $\endgroup$ – Harald Hanche-Olsen Sep 1 '16 at 14:32
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    $\begingroup$ And there I hit the character limit (had to abbreviate “point” at the end). If you need a more detailed answer, you should ask a new question, referencing this one, and explain as best you can where you get stuck. $\endgroup$ – Harald Hanche-Olsen Sep 1 '16 at 14:33
  • $\begingroup$ "Note that near any point with f′(w)≠0f′(w)≠0, ff has a local analytic inverse, so there is no branch point." Are you using the theorem that zero's of analytic functions are isolated, when you say this? $\endgroup$ – Arkya Chatterjee Sep 1 '16 at 18:36
  • $\begingroup$ Well, no, it's more elementary – just the inverse function theorem. It's been too long since I taught complex function theory, I no longer remember exactly where in the sequence this usually appears, but I'm pretty sure it happens early on. The $n$-variable holomorphic function is available here on wikipedia. $\endgroup$ – Harald Hanche-Olsen Sep 1 '16 at 20:24

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