Zeros and poles are closely related and their relationship may be used to calculates residues. First we introduce two theorems without proof. (Their proofs can be found, for instance, in [1].)

*Theorem* 1. A function $f$ is that is analytic at a point $z_0$ has a zero of order $m$ there if and only if there is a function $g$, which is analytic and nonzero at $z_0$, such that

$$f(z)=(z-z_0)^mg(z).$$

*Theorem* 2. Suppose that:

(i) two functions $p$ and $q$ are analytic at a point $z_0$;

(ii) $p(z_0)\ne 0$ and $q$ has a zero of order $m$ at $z_0$.

Then $\frac{p(z)}{q(z)}$ has a pole of order $m$ at $z_0$.

Now we discuss our main theorem in this lecture.

*Theorem*. Let two functions $p$ and $q$ be analytic at a point $z_0$. If

$$p(z_0)\ne 0,\ q(z_0)=0,\ \mbox{and}\ q’(z_0)\ne 0,$$

then $z_0$ is a simple pole of $\frac{p(z)}{q(z)}$ and

$$\mathrm{Res}_{z=z_0}\frac{p(z)}{q(z)}=\frac{p(z_0)}{q’(z_0)}.$$

*Proof*. From the conditions, we see that $q(z)$ has a zero of order $1$, so by theorem 1 it can be written as

$$q(z)=(z-z_0)g(z)$$

where $g(z)$ is analytic at $z=z_0$ and $g(z_0)\ne 0$. This can be in fact readily seen without quoting theorem 1. Since $q(z)$ is analytic at $z_0$, it can be written as a Taylor series expansion

\begin{align*}

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

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

&=(z-z_0)\left\{\frac{q’(z_0)}{1!}+\frac{q^{\prime\prime}(z_0)}{2!}(z-z_0)+\cdots\right\}.

\end{align*}

Set $g(z)=\frac{q’(z_0)}{1!}+\frac{q^{\prime\prime}(z_0)}{2!}(z-z_0)+\cdots$. Then $g(z)$ is analytic at $z=z_0$ and that $g(z_0)=q’(z_0)\ne 0$.

Now, theorem 2 implies that $\frac{p(z)}{q(z)}$ has a simple pole at $z=z_0$, but without quoting theorem 2, $\frac{p(z)}{q(z)}=\frac{\frac{p(z)}{g(z)}}{z-z_0}$, $\frac{p(z)}{g(z)}$ is analytic and nonzero at $z=z_0$. So, $\frac{p(z)}{q(z)}$ has a simple pole at $z=z_0$ as we studied here. Thus,

$$\mathrm{Res}_{z=z_0}\frac{p(z)}{q(z)}=\frac{p(z_0)}{g(z_0)}=\frac{p(z_0)}{q’(z_0)}.$$

This completes the proof.

*Example*. Find the residue of the functions

$$f(z)=\frac{z}{z^4+4}$$

at the isolated singularity $z_0=\sqrt{2}e^{\frac{i\pi}{4}}=1+i$.

*Solution*. Let $p(z)=z$ and $q(z)=z^4+4$. Then $p(z_0)=p(1+i)=1+i\ne 0$, $q(z_0)=0$, and $q’(z_0)=4z_0^3=4(1+i)^3\ne 0$. Hence, $f(z)$ has a simple pole at $z_0$. The residue $B_)$ is found by

$$B_0=\mathrm{Res}_{z=z_0}f(z)=\frac{p(z_0)}{q’(z_0)}=\frac{z_0}{3z_0^3}=\frac{1}{4z_0^2}=-\frac{i}{8}.$$

*Example*. Evaluate the contour integral

$$\int_C\frac{e^{zt}}{\sinh z}dz,$$

where $C$ is the positively oriented circle $|z|=8$.

*Solution*. $\sinh z=0$ if and only if $e^{2z}=1$. Let $z=x+iy$. Then it follows from $e^{2z}=1$ that $$e^{2x}\cos 2y=1,\ e^{2x}\sin 2y=0.$$

The solutions are $x=0$ and $y=n\pi$, $n=0,\pm 1,\pm 2,\cdots$. Thus, $\sinh z=0$ if and only if $z=n\pi i$, $ n=0,\pm 1,\pm 2$. Among these, only $z_0=0$, $z_1=\pi i$, $z_2=-\pi i$, $z_3=2\pi i$, and $z_4=-2\pi i$ lie within the interior of the circle $|z|=8$. Let $p(z)=e^{zt}$ and $q(z)=\sinh z$. Then for each $i=0,1,2,3,4$, $p(z_i)\ne 0$, $q(z_i)=0$ and $q(z_i)\ne 0$. So each $z_i$ is a simple pole of $f(z)=\frac{e^{zt}}{\sinh z}$. If we denote each residue $\mathrm{Res}_{z=z_i}f(z)$ by $B_i$, then we have

\begin{align*}B_0&=\frac{p(0)}{q’(0)}=1,\\

B_1&=\frac{p(\pi i)}{q’(\pi i)}=-e^{\pi it},\ B_2=\frac{p(-\pi i)}{q’(-\pi i)}=-e^{-\pi it},\\

B_3&=\frac{p(2\pi i)}{q’(2\pi i)}=e^{2\pi it},\ B_4=\frac{p(-2\pi i)}{q’(-2\pi i)}=e^{-2\pi it}.\end{align*}

By Cauchy’s Residue Theorem, we obtain

\begin{align*}

\int_C\frac{e^{zt}}{\sinh z}dz&=2\pi i\sum_{i=0}^4B_i\\

&=2\pi i(1-2\cos\pi t+2\cos 2\pi t).

\end{align*}

*References*:

[1] James Brown and Ruel Churchill, Complex Variables and Applications, 8th Edition, McGraw-Hill, 2008