Given that $z = 0$ and $z = 2$ are simple poles of $f$, and $\int_{|z| = 1} f(z) dz = 1$ and $\int_{|z| = 3} f(z) dz = 0$, find $f$

Question

The function $f$ is analytic everywhere on the extended complex plane except at $z = 0$ and $z = 2$, which are simple poles of $f$. Suppose $\int_{|z| = 1} f(z) dz = 1$ and $\int_{|z| = 3} f(z) dz = 0$. Find $f$, up to an arbitrary constant.

I postulate that $f(z) = \frac{g(z)}{z(z - 2)}$, with $g$ being an everywhere-analytic function. The integrals can be used to show that $g(0) = g(2) = \frac{i}{\pi}$, since $\text{Res}(f, 0) = \frac{g(0)}{0 - 2} = -\frac{1}{2}g(0)$, $\int_{|z| = 1} f(z) dz = 2\pi i \text{Res}(f, 0) = 1$, and $\int_{|z| = 3} f(z) dz = 2\pi i (\text{Res}(f, 0) + \text{Res}(f, 2)) = 0 \iff \text{Res}(f, 0) = -\text{Res}(f, 2)$. Together these suggest the value of $g(0)$ and $g(2)$, after some algebra.

I believe that $g$ is a constant function (that is, $g(z) = \frac{i}{\pi}$) and thus that $f(z) = \frac{i}{\pi z (z - 2)} + C$, but I'm not sure how to show this. I think it involves some result of the Liouville theorem, that says that entire functions with a finite number of simple poles are rational, but I'm unsure about this. What do I need to do to wrap this up?

Answer 1

Subtracting a function with two poles gives you something analytic on the extended complex plane. In particular it is entire and bounded, so Liouville says it is constant.

Answer 2

Every anaytic function whose domain is the extended complex plane is constant. Apply this statement to $g$.