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

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

Because $g(z)$ 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

$$\int_{C}\frac{1}{(z-a)^{n}}\; dz = \int_{0}^{2\pi}\frac{i\varepsilon e^{i\theta}\;d\theta}{\varepsilon^{n}e^{i\theta}}$$

$$= \frac{i}{e^{n-1}}\int_{0}^{2\pi}e^{(1-n)i\theta}\;d\theta$$

$$= \left\{\begin{array}{l} i\theta\mid_{0}^{2\pi}\\ \frac{i}{e^{n-1}}\frac{e^{(1-n)i\theta}}{(1-n)i}\mid_{0}^{2\pi} \end{array}\right.$$

$$= \left\{\begin{array}{l} 2\pi i\\ 0 \end{array}\right.$$

Exercise5.2

1. Find the residue at the singularity of the following function.

(a)

$\frac{1}{z(z-1)^2}$

(b)

$\frac{z}{(2z+1)(z-2)}$

(c)

$\frac{1}{\sin{z}}$

(d)

$\frac{e^{z}}{(z-1)(z+2)^2}$

2. Evaluate the following integrals.

(a)

$\int_{\vert z\vert=2}\frac{dz}{z(z-1)^2}$

(b)

$\int_{\vert z\vert=1}\frac{z}{(2z+1)(z-2)} dz$

(c)

$\int_{\vert z\vert=1}\frac{dz}{\sin{z}}$

(d)

$\int_{\vert z\vert=3}\frac{e^{z}}{(z-1)(z+2)^2} dz$

3. Evaluate $\frac{e^{2z}}{z^{2}(z^2 + 2z + 2)}$ along the following curves.

(a)

$\vert z\vert = 1$

(b)

$\vert z - i\vert = 2$

(c)

$\vert z\vert = 3$