Does $f(z)=\frac{2z-1}{z^2(z-1)^2}$ have an antiderivative in given domain.

Question

I have the following problem I'm having trouble with :

Question :

Let $f(z)=\frac{2z-1}{z^2(z-1)^2}$ find the domain of $f$ : $D_f$ , does $f$ have an antiderivative in $D_f$

Attempts :

$D_f= \mathbb{C} $ \ {$0,1$}

First thing I did was rewriting $f(z)$ using partial fractions I got :

$f(z)=\frac{2z-1}{z^2(z-1)^2} = \frac{1}{(z-1)^2}- \frac{1}{z^2}$ , but I got nothing( I thought using the linearity rules of integration proving that both $\frac{1}{z^2}$ and $\frac{1}{(z-1)^2}$ have one )

Or maybe should I use the residue theorem : $\int_\gamma f(z)~\mathrm dz=2\pi i \sum_{k=1}^n\operatorname{Res}( f, z_k )\,\mathrm{Ind}_\gamma(z_k).$ ?

Any hint or help would be a lot appreciated , thanks in advance

Answer 1

It is immediately obvious that $\frac{1}{(z-1)^2} - \frac{1}{z^2}$ has antiderivative $\frac{1}{z} - \frac{1}{z-1}$, which is well defined on $D_f$. Lets say it isnt - the residue theorem tells us (since the residue of f around both of its’ poles is 0) that the integral of f on any closed curve in $D_f$ is 0 - which means f has an antiderivative in $D_f$