Residues.

Let $ f$ be a function analytic on a simple Jordan curve $ C$ and at all the interior points, excepted at $ z_0$ . The residue of $ f$ at $ z_0$ , denoted Res[$ f(z),z_0$ ] is the complex number;

Res$\displaystyle [f(z),z_0] = \frac {1}{2 \pi i} \int_C f(z) \; dz$    

Theorem 9.1.1   The residue of $ f$ at $ z_0$ is the coefficient $ a_{-1}$ in the Laurent series expansion of $ f$ in an annulus $ 0<\vert z-z_0\vert<r$ .

Example 9.1.2   Compute the integral $ \int_C \frac {1}{z^2+z} \; dz$ , where $ C=\{ z \in \mathbb{C}; \; \vert z+1\vert=1 \}$ .

We compute a Laurent series expansion for $ z$ which is convergent on an annulus centered at -1.

$\displaystyle f(z)$ $\displaystyle = \frac {1}{z^2+z} =\frac { 1}{z(z+1)}$    
$\displaystyle \quad$ $\displaystyle = \frac {1}{z+1} \cdot \frac {1}{(z+1)-1}$    
$\displaystyle \quad$ $\displaystyle = - (z+1)^{-1} \cdot \frac {1}{1-(z+1)}$    
$\displaystyle \quad$ $\displaystyle = - (z+1)^{-1} ( 1 + (z+1) + (z+1)^2 + \dots )$    
$\displaystyle \quad$ $\displaystyle = - (z+1)^{-1} -1 - (z+1) - (z+1)^2 - \dots$    

Therefore $ \int_C \frac {1}{z^2+z} \; dz = 2 \pi i (-1) = -2 \pi i$ .

How to compute quickly residues?
  1. If $ f$ has a pole of order 1 at $ z_0$ , then

    Res$\displaystyle [f(z),z_0]= \underset{z \rightarrow z_0}{\text{lim}} (z-z_0)f(z).$    

  2. If $ f$ has a pole of order 2 at $ z_0$ , then

    Res$\displaystyle [f(z),z_0]= \underset{z \rightarrow z_0}{\text{lim}} \frac {d}{dz}((z-z_0)^2f(z)).$    

  3. If $ f$ has a pole of order n at $ z_0$ , then

    Res$\displaystyle [f(z),z_0]= \underset{z \rightarrow z_0}{\text{lim}} \frac {1}{(n-1)!} \frac {d^{n-1}}{dz^{n-1}}((z-z_0)^nf(z)).$    

  4. If $ f(z)=g(z)/h(z)$ has a simple pole at $ z_0$ , then

    Res$\displaystyle [f(z),z_0]= \frac {g(z_0)}{h'(z_0)}$    

Noah Dana-Picard 2007-12-24