# Existence of meromorphic root for meromorphic function

The exercises states:

Let $f:D \to \mathbb{C}$ be a meromorphic function with a finite number of roots and poles, $D$ being some simply-connected open set.

Prove that there exists a meromorphic function $g:D \to \mathbb{C}$ such that $g^2=f$ in $D$, if and only if every zero and every pole of $f$ are of even degree.

($g^2$ means $g$ multiplied by $g$ (nothing else)).

Now, the following question "Continuous root for analytic functions" deals with a similar case, but it is stated there that $f$ has only one zero (and no poles).

How to prove when there are multiple zeros? And poles?

Thanks for the help!

• Did you try the same proof for your problem?
– user301452
Commented Feb 9, 2018 at 6:59
• @PaulK I think the proof works well for single zero / pole, but not to a combination of them. If there are multiple zeros / poles, I can find a root for each one of them in a neighborhood where the zero / pole is the single special point, but not to the whole domain. Commented Feb 9, 2018 at 7:12

If such a $$g$$ exists, then, for each $$a\in D$$, if $$g$$ has a zero of order $$n$$ at $$a$$, then $$f$$ has a zero of order $$2n$$ at $$a$$ and if $$g$$ has a pole of order $$n$$ at $$a$$, then $$f$$ has a pole of order $$2n$$ at $$a$$.
On the other hand, if all zeros and all poles of $$f$$ have even order, let $$P$$ be the set of poles and let $$Z$$ the set of zeros. For each $$a\in P\cup Z$$, let $$o(a)$$ be its order. Then the function$$z\mapsto\frac{\prod_{a\in P}(z-a)^{o(a)}}{\prod_{a\in Z}(z-a)^{o(a)}}f(z)$$only has removable singularities in $$D$$ and it can be extended to an analytic function $$h\colon D\longrightarrow\mathbb C$$ without zeros. Since $$h$$ has no zeros and $$D$$ is simply connected, $$h$$ has an analytic square root $$\psi\colon D\longrightarrow\mathbb C$$. So, define$$g(z)=\frac{\prod_{a\in Z}(z-a)^{\frac{o(a)}2}}{\prod_{a\in P}(z-a)^{\frac{o(a)}2}}\psi(z)$$and then $$g^2=f$$.
• In $g(z)$ shouldn't the numerator be over the zeros, and the denominator be over the poles?