Residue

In the Laurent expansion of $f(z)$ centered on the point $a$ , the series part is set to $g(z)$ . Then we can write

$\displaystyle f(z) = g(z) + \frac{c_{-1}}{z-a} + \frac{c_{-2}}{(z-a)^2} + \cdots + \frac{c_{-n}}{(z-a)^{n}} + \cdots$

Because

is regular when $C$

is centered on a circle of $a$

, $\int_{C}g(z)\;dx = 0$ . On the other hand, the series of the principal part converges on $C$

. Therefore, if the Laurent expansion of $f(z)$

is term-wise integrated along $C$

, then we have

$\displaystyle \int_{C}\frac{1}{(z-a)^{n}}\; dz$	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\frac{i\varepsilon e^{i\theta}\;d\theta}{\varepsilon^{n}e^{i\theta}}$
	$\displaystyle =$	$\displaystyle \frac{i}{e^{n-1}}\int_{0}^{2\pi}e^{(1-n)i\theta}\;d\theta$
	$\displaystyle =$	$\displaystyle \left\{\begin{array}{l} i\theta\mid_{0}^{2\pi}\\ \frac{i}{e^{n-1}}\frac{e^{(1-n)i\theta}}{(1-n)i}\mid_{0}^{2\pi} \end{array}\right.$
	$\displaystyle =$	$\displaystyle \left\{\begin{array}{l} 2\pi i\\ 0 \end{array}\right.$