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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!

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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 $0$.

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  • $\begingroup$ beautiful!I go into a straits to prove that the origin contained in f(z),if we argue,it seems convinent better! $\endgroup$
    – ymm
    Feb 3 at 8:28
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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)$.

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  • $\begingroup$ this looks beautiful,but can you tell me why $(z- $z_0$ )^n$ does not have a holomorphic logarithm in any disk around $z_0$ $\endgroup$
    – ymm
    Feb 3 at 8:31
  • $\begingroup$ @ymm That is very standard. Any branch of logarithnm of ths function has a discontinuity along a line segment. For example, the principal branch is discontinuous along the segment $z_0-r$ for all $r>0$ such that $z_o-r$ lies in the disk. $\endgroup$ Feb 3 at 8:35

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