# Prove meromorphic function can be written as product of holomorphic and rational function

I'm not able to prove this. Any help would be welcomed !

Let U be a simply connected domain and let $f$ be a meromorphic function on U with only finitely many zeroes and poles. Prove that there is a holomorphic function $g$ : U $\rightarrow \mathbb{C}$ and a rational function $q$, such that
$\forall z$ $\in$ U : $f(z) = e^{g(z)}q(z)$

• Start with the case of a function with neither poles nor zeros. Reduce the general case to this case afterwards. Jan 8 '14 at 11:59
• If I understand right I should first prove that if $f$ has no pole nor zeroes, it holds that $\forall z \in U$ $\exists$ $g$ holomorphic : $f(z) = e^{g(z)}$ ? Jan 8 '14 at 12:10
• Order of quantifiers is important. The statement "$\forall z \in U$, $\exists g$ holomorphic : $f(z) = e^{g(z)}$" is trivial. The statement you want is "$\exists g$ holomorphic: for all $z\in U$, $f(z) = e^{g(z)}$." Jan 8 '14 at 12:12
• Right (Except for the order of quantifiers mentioned by Willie Wong). You can probably just refer to a theorem earlier in the book or course for that. Jan 8 '14 at 12:12
• Oops thanks @WillieWong ! And I found the theorem in my notes ;-) Jan 8 '14 at 12:59

As @DanielFischer proposed we at first prove the case were $f$ has neither poles nor zeros: We want to prove that if $f$ has no pole nor zeroes, it holds that $\exists \,g:U \rightarrow\mathbb{C}$ such that $\forall \,z\in U: f(z)=e^{g(z)}$
We know that since $f$ has no zeroes in $U$ the function $h(z) = \frac{f'(z)}{f(z)}$ is well defined in $U$.
Futhermore $U$ is simply connected therefore $\exists \, F: U \rightarrow\mathbb{C}: F'=h=\frac{f'(z)}{f(z)}$
We now define $\phi : U \rightarrow\mathbb{C}$ and $\phi(z):=e^{-F(z)}f(z)$
We see that $\phi$ is constant in $U$ since $$\phi'(z)= e^{-F(z)}(f'(z)-F'(z)f(z))=e^{-F(z)}(f'(z)-h(z)f(z))=e^{-F(z)}(f'(z)-\frac{f'(z)}{f(z)}f(z))=0$$ The last little definition, which will define our searched $g$, needs that we pick a point $z_0 \in U$ and another $w_0$ (depending on $z_0$) such that $e^{w_0}=f(z_0)$
Now $$g:U\rightarrow\mathbb{C} ~~~~g(z):=F(z)-F(z_0)+w_0$$ is holomorphic since $F$ is and it holds for all $z \in U:$ $$e^{g(z)}=e^{F(z)}e^{-F(z_0)}e^{w_0}=e^{F(z)}\phi(z_0)=e^{F(z)}\phi(z)=f(z)$$
If $f$ has poles and zeroes:
Let $z_1,...,z_k$ be the poles (multiplicity $m_{z_1},...,m_{z_k}$) and $w_1,...,w_l$ (multiplicity $m_{w_1},...,m_{w_l}$) the zeroes of $f$ then we just have to write $f$ as $$f(z)=e^{g(z)}\frac{\prod_{j=1}^{l}(w-w_j)^{m_{w_j}}}{\prod_{m=1}^{k}(z-z_m)^{m_{z_m}}}$$