Monthly Archives: April 2014

Zeros and Poles

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

Residues at Poles

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

The Three Types of Isolated Singularities

Recall that if $f(z)$ has an isolated singularity at $z=z_0$, it may be represented by a Laurent series
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_n}{(z-z_0)^n}+\cdots$$
in a puctured disk $0<|z-z_0|<R$. The part of series that contains negative powers of $z-z_0$
$$\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_n}{(z-z_0)^n}+\cdots$$
is called the principal part of $f(z)$ at $z_0$.

Suppose that there exists $m\in\mathbb{N}$ such that $b_m\ne 0$ and $b_{m+1}=b_{m+2}=\cdots=0$. Then
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_m}{(z-z_0)^m},$$
where $0<|z-z_0|<R$. In this case, the isolated singularity $z_0$ is called a pole of order $m$. A pole of order 1 is usually called a simple pole.

Example.
\begin{align*}
\frac{z^2-2z+3}{z-2}&=z+\frac{3}{z-2}\\
&=2+(z-2)+\frac{3}{z-2},\ 0<|z-2|<\infty
\end{align*}
has a simple pole at $z_0=2$. The residue at $z_0=2$ is 3.

Example.
\begin{align*}
\frac{\sinh z}{z^4}&=\frac{1}{z^4}\left(z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots\right)\\
&=\frac{1}{z^3}+\frac{1}{3!z}+\frac{z}{5!}+\frac{z^3}{7!}+\cdots,\ 0<|z|<\infty
\end{align*}
has a pole of order $m=3$ at $z_0=0$. The residue at $z_0=0$ is $\frac{1}{6}$.

When $b_n=0$ for all $n\geq 1$, so that
$$f(z)=\sum_{n=0}^\infty (z-z_0)^n$$
the point $z_0$ is called a removable singularity.

Example.
\begin{align*}
f(z)&=\frac{1-\cos z}{z^2}\\
&=\frac{1}{z^2}\left[1-\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+\cdots\right)\right]\\
&=\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}-\cdots,\ 0<|z|<\infty.
\end{align*}
Thus $z_0=0$ is a removable singularity. Define
$$g(z)=\left\{\begin{array}{ccc}
f(z) & \mbox{if} & z\ne 0,\\
\frac{1}{2!} & \mbox{if} & z=0.
\end{array}\right.$$
Then $g(z)$ is entire.

When an infinite number of the $b_n$ are nonzero, $z_0$ is called an essential singularity.

Example.
\begin{align*}
\exp\left(\frac{1}{z}\right)&=\sum_{n=0}^\infty\frac{1}{n! z^n}\\
&=1+\frac{1}{1!z}+\frac{1}{2!z^2}+\cdots,\ 0<|z|<\infty
\end{align*}
has an essential singularity at $z_0=0$.

Example. $e^z=-1$ when $z=(2n+1)\pi i$ $(n=0,\pm 1, \pm 2,\cdots)$. So $e^{\frac{1}{z}}=-1$ when $\frac{1}{z}=(2n+1)\pi i$ or $z=-\frac{i}{(2n+1)\pi}$ $(n=0,\pm 1,\pm 2,\cdots)$and an infinite number of these points lie in any neighbourhood of the essential singularity $z_0=0$.

Picard’s Theorem. In each neighbourhood of an essential singularity, a function assumes every finite value, with one possible exception, an infinite number of time.

In the above example, since $e^{\frac{1}{z}}\ne 0$ for all $z$, $z=0$ is the exceptional value in Picard’s theorem.

More on Residues

Here and here, we studied how to evaluate the contour integral $\oint_C f(z)dz$ when $f(z)$ is analytic everywhere within and on the positively oriented simple closed contour $C$ except for a finite number of isolated singularities interior to $C$. The calculation of residues however can be a pain if there are many isolated singularities of $f(z)$ interior to $C$. It turns out that by slightly modifying the function, we may just need to deal with only one isolated singularity regardless of how many isolated singularities of $f(z)$ there are interior to $C$. This gives a great advantage from computational viewpoint.

Theorem. If a function $f$ is analytic everywhere except for a finite number of isolated singularities interior to a positively oriented simple closed contour $C$, then
$$\oint_C f(z)dz=2\pi i\mathrm{Res}_{z=0}\left[\frac{1}{z^2}f\left(\frac{1}{z}\right)\right].$$

Proof.

From the above picture, we see that the function $f(z)$ has a Laurent series expansion
$$f(z)=\sum_{n=-\infty}^\infty c_n z_n\ (R_1<|z|<\infty)$$
where
$c_n=\frac{1}{2\pi}\oint_{C_0}\frac{f(z)}{z^{n+1}}dz$ $(n=0,\pm 1,\pm 2,\cdots)$. In particular, we have
$$\oint_{C_0}f(z)dz=2\pi ic_{-1}.$$
Since the condition of validity with the representation is not of the type $0<|z|<R_2$, $c_{-1}$ is not the residue of $f$ at $z=0$. Let us replace $z$ by $\frac{1}{z}$ in the representation. Then
$$\frac{1}{z^2}f\left(\frac{1}{z}\right)=\sum_{n=-\infty}^\infty\frac{c_n}{z^{n+2}}=\sum_{n=-\infty}^\infty\frac{c_{n-2}}{z^n}\ \left(0<|z|<\frac{1}{R_1}\right).$$
Hence,
$$c_{-1}=\mathrm{Res}_{z=0}\left[\frac{1}{z^2}f\left(\frac{1}{z}\right)\right]$$
and
$$\int_{C_0}f(z)dz=2\pi i\mathrm{Res}_{z=0}\left[\frac{1}{z^2}f\left(\frac{1}{z}\right)\right].$$
Since $f$ is analytic throughout the region bounded by $C$ and $C_0$ (topologically speaking $C_0$ is homotopic to $C$),
$$\oint_C f(z)dz=\oint_{C_0}f(z)dz.$$

Example. Evaluate $\oint_C\frac{5z-2}{z(z-1)}dz$ where $C:\ |z|=2$.

Solution. Let $f(z)=\frac{5z-2}{z(z-1)}$. Then
\begin{align*}
\frac{1}{z^2}f\left(\frac{1}{z}\right)&=\frac{5-2z}{z(1-z)}\\
&=\frac{5-2z}{z}\cdot\frac{1}{1-z}\\
&=\left(\frac{5}{z}-2\right)(1+z+z^2+\cdots)\\
&=\frac{5}{z}+3+3z+\cdots\ (0<|z|<1).
\end{align*}
Thus,
$$\oint_C\frac{5z-2}{z(z-1)}dz=2\pi i(5)=10\pi i.$$

Cauchy’s Residue Theorem

In here, we discussed that if a function $f(z)$ is analytic except at an isolated singularity $z_0$ interior to a positively oriented simple closed contour $C$, then
$$\oint_C f(z)dz=2\pi i\mathrm{Res}_{z=z_0}f(z).$$
What if there are more than one isolated singularities of $f(z)$ interior to $C$? It turns out that:

Theorem [Cauchy's Residue Theorem]. Let $C$ be a simple closed contour, positively oriented. If a function $f(z)$ is analytic except inside and on $C$ except for a finite number of singularities $z_k$ $(k=1,2,\cdots)$ inside $C$, then
$$\oint_C f(z)dz=2\pi i\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z).$$

Proof.

By Cauchy-Goursat Theorem, we have
$$\oint_C f(z)dz-\sum_{k=1}^n\oint_{C_k}f(z)dz=0$$
and so,
\begin{align*}
\oint_C f(z)dz&=\sum_{k=1}^n\oint_{C_k}f(z)dz\\
&=\sum_{k=1}^n\mathrm{Res}_{z=z_k}f(z).
\end{align*}

Example. Evaluate $\oint_C\frac{5z-2}{z(z-1)}dz$, where $C$ is the circle $|z|=2$.

Solution.

For the punctured disk $0<|z|<1$,
\begin{align*}
\frac{5z-2}{z(z-1)}&=\frac{5z-2}{z}\cdot\frac{-1}{1-z}\\
&=\frac{5z-2}{z}(-\sum_{n=0}^\infty z^n)\\
&=-\left(5-\frac{2}{z}\right)\sum_{n=0}^\infty z^n\\
&=-5\sum_{n=0}^\infty z^n+2\sum_{n=0}^\infty z^{n-1}.
\end{align*}
Hence, the residue is $b_1=\mathrm{Res}_{z=0}f(z)=2$, where $f(z)=\frac{5z-2}{z(z-1)}$.

For the punctured disk $0<|z-1|<1$,
\begin{align*}
\frac{5z-2}{z(z-1)}&=\frac{5(z-1)+3}{z-1}\cdot\frac{1}{1+(z-1)}\\
&=\left(5+\frac{3}{z-1}\right)\sum_{n=0}^\infty (-1)^n(z-1)^n.
\end{align*}
Hence, the residue is $b_2=\mathrm{Res}_{z=1}f(z)=3$.
Therefore, by Cauchy’s Residue Theorem, we obtain
$$\oint_C\frac{5z-2}{z(z-1)}dz=2\pi i(b_1+b_2)=10\pi i.$$