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

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

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 $b_1=0$, $\oint_C e^{\frac{1}{z^2}}dz=0$.

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*}

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 $\cos z$
$$\cos z=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}.$$