Existence of holomorphic Antiderivative of$f(z)=\frac{z^2}{z^2-1}, \quad f: \{z\in \mathbb{C}: |z| >2\} \to \mathbb{C}$

Question

I am supposed to show that for $f(z)=\frac{z^2}{z^2-1}, \quad f: \mathbb{C}\{-1,1\} \to \mathbb{C}$ the restriction of f on $U := \{z\in \mathbb{C}: |z| >2\} $ has a holomorphic Antiderivative.
First of all I know, that U is open and simply connected - it's the complexe plane "with a whole of radius 2". My idea was to somehow use Cauchy's Theorem on this one, but what I can't check every possible path. How do I work this out? Furthermore, I already figured out, that the index of every path, that doesn't go around the hole is zero.

Answer 1

You probably mean Morera's theorem. Cauchy is the statement that if it is holomorphic, then any integral over a closed loop is $0$. Morera's theorem is essentially the converse.

Why can't you compute the integral for every closed curve? You should be able to at this point of your complex analysis course. In fact, this is the most appropriate way (and currently the only way I know how) to show that $f$ is holomorphic.

Hint: Do partial fraction decomposition and then compute the integrals seperately. What's the result?

Answer 2

As Sellerie already pointed out, you can compute $\int \frac{z^2}{z^2-1} dz$ and then you see that the result will be analytic on $U$. Actually, if you use the principal branch of the logarithm, the antiderivative will be analytic on $V := \{ z \in \mathbf{C} : |z| > 1\}$.