1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $w=\pm1$ ($z=0$).

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.