# Proving a function $f$ has a holomorphic logarithm $g$

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 $$\lambda$$?

• True, sorry. I'll edit it now – codingnight Jun 15 at 17:35
• 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 $2\pi i(A+B)$. – David C. Ullrich Jun 15 at 17:35
• $z/(z-1)$ has a logarithm (because the $+2i\pi,-2i\pi$ in $\log z - \log (z-1)$ cancel each other) but not $z, z-1,z(z-1)$ – reuns Jun 15 at 17:52

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}
• 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? – codingnight Jun 15 at 17:41
• We are talking about a function whose domain is $\mathbb C\setminus[0,1]$ here. And $e^{i\times0}=1\notin\mathbb C\setminus[0,1]$, – José Carlos Santos Jun 15 at 17:43