While reading through various notes dealing with holomorphic function, I came across this (seemingly innocent) question that caught my eye: If $f(z)$ is holomorphic in a domain $D \subset \mathbb{C}$, must $\sqrt{f(z)}$ have a branch point at every zero of f?

Can anyone provide some insight to this question?

Thank you!


No. Let $z_0$ be a zero of $f$ in $D$; wlog we may assume that $z_0 = 0$ (by making the change of variable $z \mapsto z + z_0,$ which doesn't change anything in the question).

Then $f$ has a local power series expansion of the form $z^m(a_0 + a_1 z + a_2 z^2 + \cdots),$ where $m$ is the order of the zero, so that $a_0 \neq 0$. Then $\sqrt{f(z)}$ has the local expansion $\sqrt{f(z)} = z^{m/2}(a_0 + a_1 z + a_2 z^2 + \cdots)^{1/2}.$ Now the binomial theorem shows that $(a_0 + a_1 z + a_2 z^2 + \cdots)^{1/2}$ is a well-defined holomorphic function in a n.h. of $0$ (once we fix a choice of $a_0^{1/2}$), and so $\sqrt{f(z)}$ has a branch point at $0$ if and only if $z^{m/2}$ does, which is to say if and only if $m$ is odd.

So $\sqrt{f(z)}$ will be branched at zeroes of odd order, but not at zeroes of even order.

  • 2
    $\begingroup$ Maria: "Can anyone provide some insight to this question?" -> Matt E.: "No." :-) $\endgroup$ – draks ... May 31 '12 at 5:38

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.