1. ## Contour Integration

Here's the problem:

The function f(z) is holomorphic everywhere on a closed contour $\Gamma$, and also within $\Gamma$ except at a finite set of points where f has poles.

(Recall that a pole of order n at $z = \alpha$ occurs where f(z) is of the form $\frac{h(z)}{(z - \alpha)^n}$, where h(z) is regular at $\alpha$).

Show that:

$\oint_r f(z)dz = 2\pi i*\{\text{sum of the residues at these poles}\}$,

where the residue at the pole $\alpha$ is $\frac{h^{(n-1)}(\alpha)}{(n-1)!}$.

Any and all help is appreciated.

2. $f(z)$ is holomorphic everywhere except its poles. And the contour $\Gamma$ doesn't have any singularities / poles on it.

Now, think of the Laurent Series of f(z) on the point a.

$f(z) = \sum_{n= -\infty}^{\infty} a_n (z-a)^n$

$f(z) = ... + a_{-2} (z-a)^{-2} + a_{-1} (z-a)^{-1} + a_0 (z-a)^{0} + a_1 (z-a)^{1} + ...$

Integrate both sides by the contour $\Gamma$,

$\oint_{\Gamma} f(z)~dz = ... ~+~ \oint_{\Gamma} a_{-2} (z-a)^{-2}~dz ~+~ \color{red}\oint_{\Gamma} a_{-1} (z-a)^{-1}~dz \color{black}~+ ~\oint_{\Gamma} a_0 (z-a)^{0}~dz ~+ ~...$

Cauchy's integral theorem tell us that $\oint_{\Gamma} (z-a)^n~dz = 0$ for any n except -1.

So it means all integrals except the red one will vanish (because they are 0).

Now we have,
$\oint_{\Gamma} f(z)~dz =a_{-1} \oint_{\Gamma} (z-a)^{-1}~dz$

Let $\Gamma$ be a circle of radius 1 with center $a$ in the complex plane.

$z = e^{it}+a$

This will give,

$\oint_{\Gamma} f(z)~dz =a_{-1} 2\pi i$

The -1st coefficient of Laurent Series of f(z) at a point is called the Residue of that function at that point.

So we can generalize this for contours with multiple poles,

$\oint_{\Gamma} f(z)~dz = 2\pi i \cdot \sum \text{Res}_{f(z)=...}$

3. Good job wingless, proud of you.

It depends how formal you want to see a proof. If you are okay with a simple geometric proof by induction you can read this. It seems to me you are satisfied with that. I can provide a more formal proof if you want to see it?

4. Oh yeah. I'd love to see the rigorous proof. The proof wingless provided was sufficient but I'm never against extras.