Application to integrals.

Theorem 9.2.1 (Residue Theorem)   Let $ f$ be a function, analytic at all the interior points of a Jordan curve $ C$ , excepted a finite number of isolated singular points $ z_k, k=1, \dots , r$ . Then:

$\displaystyle \int_C f(z) \; dz = 2 \pi i \underset{k=1}{{\overset{r}{\sum}}}$   Res$\displaystyle (f, z_k)$    

Note that:

Example 9.2.2   We wish to compute the integral $ \int_{\vert z\vert=2} f(z) \; dz$ , where

$\displaystyle f(z)= \frac {e^z}{(z-1)^2(z-i)(z-3)^3}.$    

The function has a pole of order 1 at $ i$ , a pole of order 2 at1 and a pole of order 3 at 3; this last pole is irrelevant to our computation as it lies out of the Jordan curve defined by $ \vert z\vert=2$ . So we compute two residues:


$\displaystyle \int_{\vert z\vert=2} f(z) = 2 \pi i \left( 10^{-3}e^i(13-9i) + \frac{e}{8}(2+i) \right)$    

Example 9.2.3   Compute the integral $ I= \int_{0}^{2 \pi} \frac {\sin \theta}{3-2 \sin \theta} \; d \theta$ .

We use 2.1: $ \sin \theta = \frac {1}{2i} (e^{i \theta} - e^{-i \theta})$ . If $ z= e^{i \theta}$ , we have

$\displaystyle I$ $\displaystyle = \int_{\vert z\vert=1} \frac {\frac {1}{2i} (z-z^{-1})}{ 3 - 2 \frac {1}{2i} (z - z^{-1})} \; dz$    
$\displaystyle \quad$ $\displaystyle = \int_{\vert z\vert=1} \frac {z-z^{-1}}{6i - 2 (z-z^{-1}} \; dz$    
$\displaystyle \quad$ $\displaystyle = \int_{\vert z\vert=1} \frac {z^2-1}{-2z^2 +6i z + 2} \; dz$    

The denominator has two complex roots $ z_1=i\frac {\sqrt{5}+3}{2}$ and $ z_2-i\frac {\sqrt{5}-3}{2}$ . Only the second one is a point interior to the unit-circle, i.e. it is the only pole of the function we have to deal with. Therefore:

$\displaystyle I$ $\displaystyle = 2 \pi i$   Res$\displaystyle [f(z), z_2]$    
$\displaystyle \quad$ $\displaystyle = 2 \pi i \underset{z \rightarrow z_2}{\text{lim}} (z-z_2)f(z)$    
$\displaystyle \quad$ $\displaystyle = 2 \pi i \underset{z \rightarrow z_2}{\text{lim}} \frac {z^2-1}{-2 \left(z+ \frac {\sqrt{5}-3}{2} i \right) }$    
$\displaystyle \quad$ $\displaystyle = \frac {3\sqrt{5}-5}{5} \pi.$    

Noah Dana-Picard 2007-12-24