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Let $f(z)=z(z-1)$, $z\in\Omega=\mathbb{C}-[0,1]$. I want to prove that $f$ does not have a holomorphic logarithm, meaning a holomorphic function $g$ such that $e^g=f$ does not exist.

For $f$ to have a holomorphic logarithm is the same as $df/f=f'(z)dz/f(z)$ having a primitive function $F$. I know that a function $f$ has a primitive $F$ if and only if $\int_\lambda f(z)dz=0$ for all closed curves $\lambda$ in $\Omega$.

In my case $df/f=f'(z)dz/f(z)=\frac{2z-1}{z(z-1)}$.

How do I prove that $$ \int_\lambda \frac{2z-1}{z(z-1)}dz\neq0 $$ for some closed curve $\lambda$?

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  • $\begingroup$ True, sorry. I'll edit it now $\endgroup$ – codingnight Jun 15 at 17:35
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    $\begingroup$ It's enough to check the curve $|z|=2$. Find $A$ and $B$ so $f'/f = A/z + B/(z-1)$; then the integral is $2\pi i(A+B)$. $\endgroup$ – David C. Ullrich Jun 15 at 17:35
  • $\begingroup$ $z/(z-1)$ has a logarithm (because the $+2i\pi,-2i\pi$ in $\log z - \log (z-1)$ cancel each other) but not $z, z-1,z(z-1)$ $\endgroup$ – reuns Jun 15 at 17:52
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Take $\gamma(t)=2e^{it}$ ($t\in[0,2\pi]$). Then\begin{align}\int_\gamma\frac{2z-1}{z(z-1)}\,\mathrm dz&=2\pi i\left(\operatorname{res}_{z=1}\left(\frac{2z-1}{z(z-1)}\right)+\operatorname{res}_{z=-1}\left(\frac{2z-1}{z(z-1)}\right)\right)\\&=2\pi i(1+1)\\&\neq0.\end{align}

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  • $\begingroup$ You took $\gamma(t)=2e^{it}$ becasue if you had taken $\gamma(t)=e^{it}$ then $\gamma^*\in [0,1]$, right? So if the interior of $\gamma$ contains a part of $\Omega^c$ it does not matter at all? $\endgroup$ – codingnight Jun 15 at 17:41
  • $\begingroup$ We are talking about a function whose domain is $\mathbb C\setminus[0,1]$ here. And $e^{i\times0}=1\notin\mathbb C\setminus[0,1]$, $\endgroup$ – José Carlos Santos Jun 15 at 17:43

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