# the nessesary condition of branch of logarithm

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!

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