Example Determine the residues of the following function．

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

Answer A singular point of a function $f(z)$ is a vlaue of $z$ at which $f(z)$ fails to be analytic. ．

Residue theorem

$$\int_{\vert z-a\vert = r}\frac{1}{(z - a)^{n}}\ dz = \left\{\begin{array}{ll} 2\pi i, & n = 1\\ 0, & n \neq 1 \end{array}\right.$$

As you can see that the integration of $\frac{1}{z-a}$ is not zero. But other integrations are all zero．In this sense, the coefficient of $\frac{1}{z-a}$ is called residue and we express $Res[a]$．

Residue Formula Suppose that a function $f(z)$ has a mth-order pole at $a$. Then

$$Res[a] = \frac{1}{(m-1)!}\lim_{z \to a}\frac{d^{m-1}}{dz^{m-1}}(z -a)^{m}f(z)$$

In this problem, we use Residue formula．

$z = 0$ is a simple pole. So,

$$Res[0] = \lim_{z \to 1}(z-1) \frac{e^z}{(z-1)(z+2)^2} = \lim_{z \to 1}\frac{e^z}{(z+2)^2} = \frac{e}{9}$$

$z = -2$ is a double pole, So,

$$Res[-2] = \lim_{z \to -2}\frac{d ((z+2)^2 \frac{e^z}{(z-1)(z+2)^2})}{dz} = \lim_{z \to -2}\frac{ze^z - 2e^z}{(z-1)^2} = \frac{-4e^{-2}}{9}$$