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

Posted in Functions of a Complex Variable | Leave a comment

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.

Posted in Functions of a Complex Variable | Leave a comment

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,
$$\oint_C\frac{5z-2}{z(z-1)}dz=2\pi i(5)=10\pi i.$$

Posted in Functions of a Complex Variable | Leave a comment

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
$$\oint_C\frac{5z-2}{z(z-1)}dz=2\pi i(b_1+b_2)=10\pi i.$$

Posted in Functions of a Complex Variable | Leave a comment

Residues

Definition. A point $z_0$ is called a singular point or a singularity of a function $f$ if $f$ fails to be analytic at $z_0$ but is analytic at some point in every neighbourhood of $z_0$. A singularity is said to be isolated if in addition there is a deleted neighbourhood $0<|z-z_0|<\epsilon$ of $z_0$ throughout which $f$ is analytic.

Example. $f(z)=\frac{z+1}{z^3(z^2+1)}$ has three isolated singularities $z=0$ and $z=\pm i$.

Example. $\mathrm{Log}z=\ln r+i\Theta$ ($r>0$, $-\pi<\Theta<\pi$), the principal branch of the logarithmic function $\log z$ has a singularity at the origin but it is not isolated.


Example. $f(z)=\frac{1}{\sin\left(\frac{\pi}{z}\right)}$ has singularities $z=0$ and $z=\frac{1}{n}$ ($n=\pm 1, \pm 2, \cdots$). Each singularity except $z=0$ is isolated.

If $z_0$ is an isolated singularity of a function $f$, then there exists $R>0$ such that $f$ is analytic throughout which $0<|z-z_0|<R$, so $f(z)$ is represented by a Laurent series
\begin{align*}
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\\
&(0<|z-z_0|<R)\\
b_n&=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{-n+1}}dz\ (n=1,2,\cdots)
\end{align*}
where $C$ is any positively oriented simple closed contour around $z_0$ and lying in the punctured disk $0<|z-z_0|<R$. The coefficient $b_1$ of $\frac{1}{z-z_0}$ in the above Laurent series expansion is called the residue of $f$ at the isolated singularity $z_0$. The residue is important since from the integral formula for $b_n$ above, we find
$$\oint_C f(z)dz=2\pi i b_1.$$
That is, the contour integral $\oint_C f(z)dz$ can be evaluated using the residue $b_1$. The residue $b_1$ of $f(z)$ at $z=z_0$ is also denoted by $\mathrm{Res}_{z=z_0}f(z)$.

Example. Consider $\oint_C\frac{dz}{z(z-2)^4}$ where $C$ is the positively oriented circle $|z-2|=1$. The integrand $\frac{1}{z(z-2)^4}$ is analytic everywhere except at the isolated singularities $z=0$ and $z=2$, so it has a Laurent series representation in the punctured disk $0<|z-2|<2$.
\begin{align*}
\frac{1}{z(z-2)^4}&=\frac{1}{(z-2)^4}\cdot\frac{1}{2+(z-2)}\\
&=\frac{1}{2(z-2)^4}\cdot\frac{1}{1-\left(-\frac{z-2}{2}\right)}\\
&=\frac{1}{2(z-2)^4}\sum_{n=0}^\infty\frac{(-1)^n}{2^n}(z-2)^n\\
&=\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}(z-2)^{n-4}\ (0<|z-2|<2)
\end{align*}
The residue of $\frac{1}{z(z-2)^4}$ at $z=2$ is $-\frac{1}{16}$, hence
$$\oint_C\frac{dz}{z(z-2)^4}=2\pi i\left(-\frac{1}{16}\right)=-\frac{\pi i}{8}.$$

Example. Evaluate $\oint_C e^{\frac{1}{z^2}}dz$ where $C$ is the unit circle $|z|=1$.

Solution. $e^{\frac{1}{z^2}}$ is analytic everywhere except at $z=0$. $e^{\frac{1}{z^2}}$ has an Laurent series expansion
$$e^{\frac{1}{z^2}}=1+\frac{1}{1!z^2}+\frac{1}{2!z^4}+\cdots\ (0<|z|<\infty).$$
Since $b_1=0$, $\oint_C e^{\frac{1}{z^2}}dz=0$.

Posted in Functions of a Complex Variable | Leave a comment

Laurent Series

If a function fails to be analytic at a point $z_0$, we cannot apply Taylor’s theorem at that point. However, it may be possible to find a series representation for $f(z)$ involving both positive and negative powers of $z-z_0$.

Theorem [Laurent's Theorem]. Suppose that a function $f$ is analytic throughout an annular domain $R_1<|z-z_0|<R_2$, centered at $z_0$, and let $C$ denote any positively oriented simple closed contour around $z_0$ and lying in that domain. Then, at each point in the domain, $f(z)$ has the series representation
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\sum_{n=1}^\infty\frac{b_n}{(z-z_0)^n}\ (R_1<|z-z_0|<R_2),$$
where
\begin{align*}
a_n&=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz\ (n=0,1,2,\cdots),\\
b_n&=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{-n+1}}dz\ (n=1,2,\cdots).
\end{align*}


The expansion can be also written as
$$f(z)=\sum_{n=-\infty}^\infty c_n(z-z_0)^n\ (R_1<|z-z_0|<R_2),$$
where
$$c_n=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz\ (n=0,\pm 1,\pm 2,\cdots).$$

Example. From the Maclaurin series expansion $e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$, we obtain Laurent series expansion for $e^{\frac{1}{z}}$
$$e^{\frac{1}{z}}=\sum_{n=0}^\infty\frac{1}{n!z^n}=1+\frac{1}{1!z}+\frac{1}{2!z^2}+\cdots\ (0<|z|<\infty).$$
The coefficient $b_1$ is
$$b_1=\frac{1}{2\pi i}\oint_C e^{\frac{1}{z}}dz=1$$
for any positively oriented simple closed contour $C$ around $0$ and lying in the domain $0<|z|<\infty$. Hence, we obtain the integral
$$\oint_C e^{\frac{1}{z}}dz=2\pi i$$
for any positively oriented simple closed contour $C$ around $0$ and lying in the domain $0<|z|<\infty$.

Example. $f(z)=\frac{1}{(z-i)^2}$ is already in the form of a Laurent series, where $z_0=i$. From
$$f(z)=\sum_{n=-\infty}^\infty c_n(z-i)^n\ (0<|z-i|<\infty),$$
we find that $c_{-2}=1$ and all other coefficients are zero. Hence, we have
$$\oint_C\frac{dz}{(z-i)^{n+3}}=\left\{\begin{array}{ccc}
0 &\mbox{if}&n\ne -2,\\
2\pi i &\mbox{if}& n=-2
\end{array}
\right.$$
for any positively oriented simple closed contour $C$ around $i$ and lying in the domain $0<|z-i|<\infty$.

Example. Let $f(z)$ be the function
$$f(z)=\frac{-1}{(z-1)(z-2)}=\frac{1}{z-1}-\frac{1}{z-2}.$$
Since $f(z)$ has two singularities $z=1$ and $z=2$, we may consider the following three different domains to obtain Laurent series expansion in each domain
$$D_1: |z|<1,\ D_2: 1<|z|<2,\ D_3: |z|>2.$$


For $D_1: |z|<1$,
$$\frac{1}{z-1}=-\frac{1}{1-z}=-\sum_{n=0}^\infty z^n$$
and
$$-\frac{1}{z-2}=\frac{1}{2-z}=\frac{1}{2}\frac{1}{1-\frac{z}{2}}.$$
Since $|z|<1$, $|z|<2$ and so $\frac{1}{1-\frac{z}{2}}=\sum_{n=0}^\infty\left(\frac{z}{2}\right)^n$. Hence, we obtain the Taylor series expansion
\begin{align*}
f(z)&=-\sum_{n=0}^\infty z^n+\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}\\
&=\sum_{n=0}^\infty(2^{-n-1}-1)z^n\ (|z|<1).
\end{align*}
For $D_2: 1<|z|<2$,
$$\frac{1}{z-1}=\frac{1}{z}\frac{1}{1-\frac{1}{z}}=\frac{1}{z}\sum_{n=0}^\infty\left(\frac{1}{z}\right)^n.$$
Hence, we obtain the Laurent series expansion
\begin{align*}
f(z)&=\frac{1}{z}\sum_{n=0}^\infty\left(\frac{1}{z}\right)^n+\frac{1}{2}\sum_{n=0}^\infty\left(\frac{z}{2}\right)^n\\
&=\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}+\sum_{n=1}\frac{1}{z^n}\ (1<|z|<2).
\end{align*}
For $D_3: |z|>2$,
\begin{align*}
f(z)&=\frac{1}{z-1}-\frac{1}{z-2}\\
&=\frac{1}{z}\sum_{n=0}^\infty\left(\frac{1}{z}\right)^n-\frac{1}{z}\sum_{n=0}^\infty\left(\frac{2}{z}\right)^n\\
&=\sum_{n=1}^\infty\frac{1-2^{n-1}}{z^n}\ (2<|z|<\infty).
\end{align*}

Posted in Functions of a Complex Variable | Leave a comment

Taylor Series

Theorem. Suppose that a function $f$ is analytic throughout a disk $|z-z_0|<R_0$ centered at $z_0$ and with radius $R_0$. Then $f(z)$ has the power series representation
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\ (|z-z_0|<R_0),$$
where
$$a_n=\frac{f^{(n)}(z_0)}{n!}\ (n=0,1,2,\cdots).$$

Proof. First consider the case $z_0=0$ and show that
$$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^n\ (|z|<R_0).$$
This particular Taylor series is called a Maclaurin series. Let us write $|z|=r$ and let $C_0$ denote any positively oriented circle centered at the origin and with radius $r_0$, where $r<r_0<R_0$.

Since $f$ is analytic inside and on the circle $C_0$ and since $z$ is interior to $C_0$, by the Cauchy Integral Formula, we have
$$f(z)=\frac{1}{2\pi i}\oint_{C_0}\frac{f(s)}{s-z}ds.$$
Note that $\frac{1}{s-z}$ may be written as
\begin{align*}
\frac{1}{s-z}&=\frac{1}{s}\frac{1}{1-\frac{z}{s}}\\
&=\sum_{n=0}^{N-1}\frac{z^n}{s^{n+1}}+z^N\frac{1}{(s-z)s^N}.
\end{align*}
Using this, $f(z)$ may be written as
\begin{align*}
f(z)&=\frac{1}{2\pi i}\oint_{C_0}\frac{f(s)}{s-z}ds\\
&=\sum_{n=0}^{N-1}\frac{1}{2\pi i}\oint_{C_0}\frac{f(s)}{s^{n+1}}ds z^n+\frac{z^N}{2\pi i}\oint_{C_0}\frac{f(s)}{(s-z)s^N}ds\\
&=\sum_{n=0}^{N-1}\frac{f^{(n)}(0)}{n!}z^n+\rho_N(z),
\end{align*}
where
$$\rho_N(z)=\frac{z^N}{2\pi i}\oint_{C_0}\frac{f(s)}{(s-z)s^N}ds.$$
We are done if we can show that $\displaystyle\lim_{N\to\infty}\rho_N=0$. By triangle inequality, $|s-z|\geq ||s|-|z||=r_0-r$. So, we obatin
\begin{align*}
|\rho_N(z)|&\leq \frac{r^N}{2\pi}\frac{M}{(r_0-r)r_0^N}2\pi r_0\\
&=\frac{Mr_0}{r_0-r}\left(\frac{r}{r_0}\right)^N.
\end{align*}
Since $r<r_0$, $\displaystyle\lim_{N\to\infty}|\rho_N(z)|=0$ i.e. $\displaystyle\lim_{N\to\infty}\rho_N=0$. Hence, we have proved that
$$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^n.$$
Now, suppose that $f(z)$ is analytic when $|z-z_0|<R_0$. Then $g(z):=f(z+z_0)$ is analytic when $|z|=|(z+z_0)-z_0|<R_0$. So, $g(z)$ has the Maclaurin series representation
$$g(z)=\sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}z^n\ (|z|<R_0)$$
or
$$f(z+z_0)=\sum_{n=0}^\infty\frac{f^{(n)}(z_0)}{n!}z^n\ (|z|<R_0).$$
Replacing $z$ by $z-z_0$, we finally obtain the Taylor series representation
$$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(z_0)}{n!}(z-z_)^n\ (|z-z_0|<R_0).$$

Example. The function $f(z)=e^z$ is entire, so it has a Maclaurin series representation for all $z\in\mathbb{C}$.
$$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}\ (|z|<\infty).$$

Example. $\sin z$ is defined as
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}.$$
So,
\begin{align*}
\sin z&=\frac{1}{2i}\left[\sum_{n=0}^\infty\frac{(iz)^n}{n!}-\sum_{n=0}^\infty\frac{(-iz)^n}{n!}\right]\\
&=\frac{1}{2i}\sum_{n=0}^\infty[1-(-1)^n]\frac{i^nz^n}{n!}\\
&=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}\ (|z|<\infty).
\end{align*}
Using the definition $\cos z=\frac{e^{iz}+e^{-iz}}{2}$, one can similarly obtain the Maclaurin series representation for $\cos z$
$$\cos z=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}.$$

Posted in Functions of a Complex Variable | Leave a comment

Cauchy’s Inequality and Liouville’s Theorem

Suppose that a function $f$ is analytic inside and on a positively oriented circle $C_R$, centered at $z_0$ and with radius $R$. Then by Cauchy Integral Formula
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_{C_R}\frac{f(z)}{(z-z_0)^{n+1}}dz.$$
Since $C_R$ is compact (i.e. closed and bounded) and $f(z)$ is continuous, $|f(z)|$ assumes a maximum (and also a minimum) on $C_R$. Let us call the maximum value of $|f(z)|$ on $C_R$ $N_R$. On $C_R$, $|z-z_0|=R$. Hence, we have the upper bound for $|f^{(n)}(z_0)|$ as
\begin{align*}
|f^{(n)}(z_0)|&=\frac{n!}{2\pi}\left|\oint_{C_R}\frac{f(z)}{(z-z_0)^{n+1}}dz\right|\\
&\leq\frac{n!M_R}{R^n}
\end{align*}
for $n=1,2,\cdots$. This inequality is called Cauchy’s Inequality.

Using Cauchy’s Inequality, we can prove the following Liouville’s Theorem.

Theorem. If $f$ is entire and bounded in the complex plane $\mathbb{C}$, then $f(z)$ is constant.

Proof. Since $f$ is bounded, there exists $M>0$ such that $|f(z)|\leq M$ for all $z\in\mathbb{C}$. Since $f$ is entire, for any $R>0$,
$$|f’(z)|\leq\frac{M}{R}$$
by Cauchy’s Inequality. As $R\to\infty$, $\frac{M}{R}\to 0$, so we obtain $|f’(z)|=0$ for all $z\in\mathbb{C}$. This implies that $f’(z)=0$ for all $z\in\mathbb{C}$ and hence, $f(z)$ is constant.

Posted in Functions of a Complex Variable | Leave a comment

Morera’s Theorem

Cauchy’s Integral Theorem says that if a function $f(z)$ is analytic throughout some simply connected domain $D$, then for any contour $C$ in $D$, $\oint_C f(z)dz=0$. It turns out that the converse of Cauchy’s Theorem is also true, namely

Theorem. If a function $f(z)$ is continuous in a simply connected domain $D$ and $\oint_C f(z)dz=0$ for every closed contour $C$ within $D$, then $f(z)$ is analytic throughout $D$.

This theorem is called Morera’s Theorem. Let us prove the theorem.

First we show that under the assumption the line integral $\int_\gamma f(z)dz$ is independent on the path $\gamma$. Let $\gamma$ be a path from $z_1$ to $z_2$ and choose $\gamma_1$ another path from $z_1$ to $z_2$ as shown in the figure.


Then $C: \gamma\cup (-\gamma_1)$ is a positively oriented contour. So, by assumption,
\begin{align*}
0&=\oint_C f(z)dz\\
&=\int_\gamma f(z)dz+\int_{-\gamma_1}f(z)dz\\
&=\int_\gamma f(z)dz-\int_{\gamma_1}f(z)dz.
\end{align*}
That is,
$$\int_\gamma f(z)dz=\int_{\gamma_1}f(z)dz.$$
Hence, the line integral $\int_\gamma f(z)dz$ does not depend on the path $\gamma$ but only on the endpoints $z_1$ and $z_2$.

Let us define $F(z)=\int_{z_0}^z f(w)dw$. Then
\begin{align*}
F(z+\Delta z)-F(z)&=\int_{z_0}^{z+\Delta z} f(w)dw-\int_{z_0}^z f(w)dw\\
&=\int_z^{z+\Delta z} f(w)dw.
\end{align*}
Since $\int_z^{z+\Delta z}dw=\Delta z$,
$$f(z)=\frac{1}{\Delta z}\int_z^{z+\Delta z}f(z) dw$$
and so, we have
$$\frac{F(z+\Delta z)-F(z)}{\Delta z}-f(z)=\frac{1}{\Delta z}\int_z^{z+\Delta z}[f(w)-f(z)]dw.$$
Since $f$ is continuous at $z$, given $\epsilon>0$ there exists $\delta>0$ such that $|f(w)-f(z)|<\epsilon$ whenever $|w-z|<\delta$. If $z+\Delta z$ is close enough to $z$ so that $|\Delta z|<\delta$, then
$$\left|\frac{F(z+\Delta z)-F(z)}{\Delta z}-f(z)\right|<\frac{1}{|\Delta z|}\epsilon|\Delta z|=\epsilon.$$
Therefore,
$$F’(z)=\lim_{\Delta z\to 0}\frac{F(z+\Delta z)-F(z)}{\Delta z}=f(z).$$
That is, $F(z)$ analytic in $D$. This means that $f(z)$ is analytic by Cauchy Integral Formula.

Posted in Functions of a Complex Variable | Leave a comment

Cauchy Integral Formula

Suppose that $f(z)$ is analytic everywhere inside and on a simple closed contour $C$, taken in the positive sense. If $z_0$ is a point exterior to $C$, then by Cauchy’s Integral Theorem,
$$\oint_C\frac{f(z)}{z-z_0}dz=0.$$
Now, the question is what would be the value of the integral
$$\oint_C\frac{f(z)}{z-z_0}dz$$
if $z_0$ is a point interior to $C$? To answer this question, let us consider a tiny circle centerd at $z_0$ and with radius $r$, also oriented counterclockwise as shown in the figure.

Then using Cauchy’s Integral Theorem, we obatin
$$\oint_C\frac{f(z)}{z-z_0}dz-\oint_{C_1}\frac{f(z)}{z-z_0}dz=0.$$
Let $z=z_0+re^{i\theta}$. Then
\begin{align*}
\oint_{C_1}\frac{f(z)}{z-z_0}dz&=\int_0^{2\pi}\frac{f(z_0+re^{i\theta})}{re^{i\theta}}rie^{i\theta}d\theta\\
&=i\int_0^{2\pi}f(z_0+re^{i\theta})d\theta\\
&\to 2\pi if(z_0)
\end{align*}
as $r\to 0$. Therefore, we obtain
$$f(z_0)=\frac{1}{2\pi}\oint_C\frac{f(z)}{z-z_0}dz.$$
This formula is called Cauchy Integral Formula.

Example. Let $C$ be the positively oriented circle $|z|=2$. The function $f(z)=\frac{z}{9-z^2}$ is analytic within and on $C$. Since $z_0=-i$ is interior to $C$, by Cauchy Integral Formula,
\begin{align*}
\oint_C\frac{z}{(9-z^2)(z+i)}dz&=\oint_C\frac{\frac{z}{9-z^2}}{z+i}dz\\
&=2\pi\frac{-i}{9-(-i)^2}\\
&=\frac{\pi}{5}.
\end{align*}

Derivatives

Suppose that $|\Delta z_0|$ is small enough so that $z_0+\Delta z_0$ is still interior to the contour $C$. Then by Cauchy Integral Formula, we obtain
$$\frac{f(z_0+\Delta z_0)-f(z_0)}{\Delta z_0}=\frac{1}{2\pi i\Delta z_0}\left\{\oint_C\frac{f(z)}{z-z_0-\Delta z_0}dz-\oint_C\frac{f(z)}{z-z_0}dz\right\}.$$
So,
\begin{align*}
f’(z_0)&=\lim_{\Delta z_0\to 0}\frac{f(z_0+\Delta z_0)-f(z_0)}{\Delta z_0}\\
&=\lim_{\Delta z_0\to 0}\frac{1}{2\pi i\Delta z_0}\left\{\oint_C\frac{f(z)}{z-z_0-\Delta z_0}dz-\oint_C\frac{f(z)}{z-z_0}dz\right\}\\
&=\lim_{\Delta z_0\to 0}\frac{1}{2\pi i\Delta z_0}\oint_C\frac{\Delta z_0f(z)}{(z-z_0-\Delta z_0)(z-z_0)}dz\\
&=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}dz.
\end{align*}
The same process can be repeated for $f’(z_0)$ and we obtain
$$f^{\prime\prime}(z_0)=\frac{2}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^3}dz.$$
Continuing, we get
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}$$
for $n=0,1,2,\cdots$ where $f^{(0)}(z)=f(z)$. This formula may also be called Cauchy Integral Formula which includes the Cauchy Integral Formula we derived earlier.

Example. If $C$ is the positively oriented unit circle $|z|=1$ and $f(z)=\exp(2z)$, then
\begin{align*}
\oint_C\frac{\exp(2z)}{z^4}dz&=\oint_C\frac{f(z)}{(z-0)^{3+1}}\\
&=\frac{2\pi i}{3!}f^{\prime\prime\prime}(0)\\
&=\frac{8\pi i}{3}
\end{align*}
where $f(z)=\exp(2z)$.

Example. Let $z_0$ be any point interior to a positively oriented simple closed contour $C$. When $f(z)=1$, Cauchy Integral Formula tells that
$$\oint_C\frac{dz}{z-z_0}=2\pi i\ \mbox{and}\ \oint_C\frac{dz}{(z-z_0)^{n+1}}=0,\ n=1,2,\cdots.$$

Posted in Functions of a Complex Variable | Leave a comment