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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!

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up vote 10 down vote accepted

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.

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