# Meromorphic functions and residues

Last time, we discussed Laurent series, which are essentially two-way power series. They are almost as nice as holomorphic functions, but not quite. Maybe we can recoup some of the lost beauty of holomorphicity by imposing a reasonable condition.

We want to allow ourselves to have points where the function isn’t defined, but let’s limit these points. Let’s require them to be isolated points. We say that a function ${f}$ is meromorphic on an open set ${U\subseteq {\mathbb C}}$ if ${f}$ is holomorphic on ${U}$, except at some number of isolated points. We write ${f\in{\mathcal M}(U)}$. These isolated points are called poles. We obviously shouldn’t expect ${f}$ to have a power series centered at one of these poles, but it does have a Laurent series.

Let ${f\in{\mathcal M}(U)}$ and let ${z_0\in U}$ be a pole of ${f}$. Then as we saw last time, ${f}$ has a Laurent series centered at ${z_0}$:

$\displaystyle f(z)=\cdots+a_{-2}(z-z_0)^{-2}+a_{-1}(z-z_0)^{-1}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots$

Moreover, the series has inner radius of convergence 0, so this representation is valid for all ${z}$ close enough to ${z_0}$.

Now if we take a loop ${\gamma}$ in ${U}$, and integrate, if ${f}$ has no poles on the interior of ${\gamma}$, then the integral is zero. This is the Cauchy integral theorem. What if it does have a pole? We can use the generalized Cauchy integral formula we saw last time:

Theorem 1 (Residue theorem) Let ${f\in{\mathcal M}(U)}$ and let ${z_0}$ be a pole of ${f}$ in ${U}$. Expand ${f}$ as a Laurent series as above. Let ${\gamma}$ be a small counter-clockwise circle about ${z_0}$ such that the only pole in its interior is ${z_0}$. Then

$\displaystyle \displaystyle\frac{1}{2\pi i}\int_\gamma f(z)dz=a_{-1}.$

Proof:The generalized Cauchy integral formula we saw last time said that

$\displaystyle a_n=\displaystyle\frac{1}{2\pi i}\int_\gamma\frac{f(z)}{(z-z_0)^{n+1}}dz.$

Let ${n=-1}$.

$\Box$

If we take a loop that goes around several poles, the integral must be the sum of the integrals, as we can build a homotopy as shown in the images below (click to blow up).

The value ${a_{-1}}$ (for an expansion about ${z_0}$) is called the residue of ${f}$ at the pole ${z_0}$. In other words, the theorem states that the value of an integral about a contour is the sum of the residues of the poles inside.