When $f(z)$ has a pole of order $m$, we may be able to find the residue of $f(z)$ at $z_0$ without expanding $f(z)$ into a Laurent series at $z=z_0$. This gives a great computational advantage.

Suppose that $z_0$ is a pole of order $m$ of $f(z)$. Then $f(z)$ has a Laurent series expansion

$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_m}{(z-z_0)^m}\ (b_m\ne 0)$$

valid in a punctured disk $0<|z-z_0|<R$.

Define

$$\phi(z)=\left\{\begin{array}{ccc}

(z-z_0)^mf(z) & \mbox{if} & z\ne z_0,\\

b_m & \mbox{if} & z=z_0.

\end{array}\right.$$

Then $\phi(z)$ has a power series representation

$$\phi(z)=b_m+b_{m-1}(z-z_0)+\cdots+b_2(z-z_0)^{m-2}+b_1(z-z_0)^{m-1}+\sum_{n=0}^\infty(z-z_0)^{m+n},$$

for $|z-z_0|<R$. That is, $\phi(z)$ is analytic at $z=z_0$ and in the disk $|z-z_0|<R$. We find that

$$

b_1=\left\{\begin{array}{ccc}\frac{\phi^{(n-1)}(z_0)}{(m-1)!} & \mbox{if} & m\geq 2,\\

\phi(z_0) & \mbox{if} & m=1.

\end{array}\right.$$

Conversely, suppose that $f(z)$ can be written in the form

$$f(z)=\frac{\phi(z)}{(z-z_0)^m},$$

where $\phi(z)$ is analytic and nonzero at $z=z_0$. $\phi(z)$ has a Taylor series expansion at $z_0$

\begin{align*}

\phi(z)=&\phi(z_0)+\frac{\phi’(z_0)}{1!}(z-z_0)+\frac{\phi^{\prime\prime}(z_0)}{2!}(z-z_0)^2\\

&+\cdots+\frac{\phi^{(m-1)}(z_0)}{(m-1)!}(z-z_0)^{m-1}+\sum_{n=m}^\infty\frac{\phi^{(n)}(z_0)}{n!}(z-z_0)^n

\end{align*}

in some neighbourhood $|z-z_0|<\epsilon$. Since $\phi(z_0)\ne 0$, $z_0$ is a pole of order $m$ of $f(z)$. Clearly, we have

$$

\mathrm{Res}_{z=z_0}f(z)=\left\{\begin{array}{ccc}

\phi(z_0) & \mbox{if} & m=1,\\

\frac{\phi^{(n-1)}(z_0)}{(m-1)!} & \mbox{if} & m\geq 2.

\end{array}\right.$$

Therefore, we have the following theorem holds.

*Theorem*. An isolated singularity $z_0$ of a function $f$ is a pole of order $m$ if and only if $f(z)$ can be written as

$$f(z)=\frac{\phi(z)}{(z-z_0)^m},$$

where $\phi(z)$ is analytic and nonzero at $z=z_0$. Moreover,

$$

\mathrm{Res}_{z=z_0}f(z)=\left\{\begin{array}{ccc}

\phi(z_0) & \mbox{if} & m=1,\\

\frac{\phi^{(n-1)}(z_0)}{(m-1)!} & \mbox{if} & m\geq 2.

\end{array}\right.$$

*Example*. $f(z)=\frac{z+1}{z^2+9}$ has an isolated singularity at $z=3i$. $f(z)$ can be written as

$$f(z)=\frac{\frac{z+1}{z+3i}}{z-3i}.$$

Then $\phi(z)=\frac{z+1}{z+3i}$ is analytic at $z=3i$ and $\phi(3i)=\frac{3-i}{6}\ne 0$. Hence, $z=3i$ is a simple pole of $f(z)$ and

$$\mathrm{Res}_{z=3i}f(z)=\phi(3i)=\frac{3-i}{6}.$$

$z=-3i$ is also a simple pole of $f(z)$ and

$$\mathrm{Res}_{z=-3i}f(z)=\frac{3+i}{6}.$$

*Example*. Let us consider $f(z)=\frac{z^3+2z}{(z-i)^3}$. $\phi(z)=z^3+2z$ is analytic at $z=i$ and $\phi(i)=i\ne 0$. Hence, $z=i$ is a pole of order $3$ and

$$b_1=\mathrm{Res}_{z=i}f(z)=\frac{\phi^{\prime\prime}(i)}{2!}=3i.$$