# 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? – Paul K Feb 9 '18 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. – co.sine Feb 9 '18 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 extende to an analytic fonction $h\colon D\longrightarrow\mathbb C$ 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$$g(z)=\frac{\prod_{a\in P}(z-a)^{\frac{o(a)}2}}{\prod_{a\in Z}(z-a)^{\frac{o(a)}2}}\psi(z)$$and then $g^2=f$.