the nessesary condition of branch of logarithm

Question

Let $U$ be an open subset of $\mathbb{C}$.Let $z_0$ be a point in U,and suppose that f is a meromorphic function on U with a pole at $z_0$.Prove that there is no holomorphic function g:U$\setminus${$z_0$}$\rightarrow$$\mathbb{C}$,such that $e^{g(z)}$=f(z)for all z$\in$U$\setminus${$z_0$}

my idea:as we know that if a simply connected domain which does not contain the origin,it can be chosen as a branch of logarithm,however,this is just sufficient condition,and we don't know what the image of f is like,thus how to deal with?this really confuses me.thanks!

Answer 1

If $g$ exists, then $f'=g'e^g$ and therefore $\frac{f'}f=g'$. So, $\frac{f'}f$ has a primitive, and therefore the integral of $\frac{f'}f$ along any closed path is $0$. But if you take $r>0$ so small that $f$ has no zero on $D(z_0,r)\setminus\{z_0\}$, then, by the Argument Principle, $\oint_{|z-z_0|=r}\frac{f'}f$ is $-2\pi i$ times the number of poles of $f$ on $D(z_0,r)$, which is not $0$.

Answer 2

If $n$ is the order of the pole then $f(z)=(z-z_0)^{-n} h(z)$ for some holomotprhic function $h$ with no zeros in some disk $B(z_0,r)$. Since this disk is simply connected and $h$ has no zeros we can write $h(z)=e^{\phi (z)}$ for some holomorpic function $\phi$. But then $(z-z_0)^{n}$ is itself an exponential in that disk with $z_0$ removed. But $(z-z_0)^{n}$ does not have a holomorpic logarithm in any disk around $z_0$ (with $z_0$ removed)$.