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 $\lambda$?
Best Answer
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}