Lecture 19: Laurent expansions

Complex analysis, lecture 2

(April 6, 2026)

The basic idea of Taylor expansions is to write a given analytic function as a power series, i.e. a linear combinations of $z^{n}$ for nonnegative $n$ . If we instead allowed $n$ to be any integer, positive or negative, then what we would get is called a Laurent series expansion. Our main goal for today is to introduce this notion, and see how we can compute it.

When working with power series, the main geometric object is a disk centered at our point of interest $z_{0}$ , with some radius (possibly $0$ or $+\infty$ ). For Laurent series, the main object will be an annulus, which can be thought of as a ring, or a disk with both an inner and outer radius: if we fix $0\leq r<R\leq+\infty$ , the corresponding annulus centered at $z_{0}$ is the region $\{z\in\mathbb{C}:r<|z-z_{0}|<R\}$ . In the special case where $r=0$ , this is $\{0<|z-z_{0}|<R\}=\{|z-z_{0}|<R\}\setminus\{z_{0}\}$ , and is sometimes called the punctured disk of radius $R$ around $z_{0}$ .

Note that unlike for disks, any annulus centered at $z_{0}$ does not contain the point $z_{0}$ . This is important: often our functions of interest will not be analytic at $z_{0}$ .

Fixing $0\leq r<R\leq+\infty$ and an annulus $D$ of radii $r, R$ centered at $z_{0}$ , suppose that $f:D\to\mathbb{C}$ is an analytic function. Let $D_{0}=\{z:|z-z_{0}|<R\}$ and $D_{1}=\{z:|z-z_{0}|>r\}$ , so that $D=D_{0}\cap D_{1}$ .

Proposition.

With notation as above, there exist unique analytic functions $f_{0}:D_{0}\to\mathbb{C}$ , $f_{1}:D_{1}\to\mathbb{C}$ with $f_{1}$ analytic at infinity with $f_{1}(\infty)=0$ such that for $z\in D$ ,

f(z)=f_{0}(z)+f_{1}(z).

The condition that $f_{1}(\infty)=0$ isn’t strictly necessary, one could also find a decomposition with a different value of $f_{1}(\infty)$ ; but imposing this condition makes the decomposition unique. (Otherwise, we could just subtract a constant from $f_{0}$ and add it to $f_{1}$ to get a different decomposition.)

Note that if $f$ were analytic on all of $D_{0}$ , not just on $D$ , then we could simply take $f_{0}=f$ and $f_{1}=0$ . Similarly, if $f$ were analytic on $D_{1}$ and at $\infty$ , we could take $f_{1}=f-f(\infty)$ and $f_{0}=f(\infty)$ .

Proof.

First, we show that if such functions $f_{0}$ , $f_{1}$ exist, then they are unique. Suppose that there was another pair $g_{0}:D_{0}\to\mathbb{C}$ , $g_{1}:D_{1}\to\mathbb{C}$ of analytic functions with $g_{1}$ analytic at infinity and $f(z)=f_{0}(z)+f_{1}(z)=g_{0}(z)+g_{1}(z)$ for $z\in D$ .

Define a function $h:\mathbb{C}\to\mathbb{C}$ by $h(z)=f_{0}(z)-g_{0}(z)$ for $z\in D_{0}$ and $h(z)=g_{1}(z)-f_{1}(z)$ for $z\in D_{1}$ ; note that by the above equation, if $z\in D=D_{0}\cap D_{1}$ then the two definitions agree, so this is well-defined. Since the $f_{i}$ and $g_{i}$ are analytic where they are defined, $h$ is analytic everywhere. As $z\to\infty$ , since $f_{1}$ and $g_{1}$ are analytic at infinity with value $0$ , $h(\infty)=\lim_{z\to\infty}h(z)=0$ , so $h$ is bounded, hence by Liouville’s theorem $h$ is constant, and since $h(\infty)=0$ we must have $h(z)=0$ for all $z$ . Hence $f_{0}(z)=g_{0}(z)$ and $g_{1}(z)=f_{1}(z)$ , i.e. $f_{0}$ and $f_{1}$ are unique.

Now we show that such $f_{i}$ actually exist. Choose some $r<r^{\prime}<R^{\prime}<R$ . For $r^{\prime}<|z-z_{0}|<R^{\prime}$ , we can take the sub-annulus of radii $r^{\prime}$ , $R^{\prime}$ centered at $z_{0}$ , which still contains $z$ , and its boundary is the union of the circles of radii $r^{\prime}$ and $R^{\prime}$ , on which $f$ is still analytic. (We need to choose $r^{\prime}$ and $R^{\prime}$ rather than just using $r$ and $R$ to make sure that $f$ extends to the boundary.) Therefore by Cauchy’s formula

f(z)=\frac{1}{2\pi i}\int_{|z-z_{0}|=R^{\prime}}\frac{f(w)}{w-z}\,dw-\frac{1}{% 2\pi i}\int_{|z-z_{0}|=r^{\prime}}\frac{f(w)}{w-z}\,dw.

Let

f_{0}(z)=\frac{1}{2\pi i}\int_{|z-z_{0}|=R^{\prime}}\frac{f(w)}{w-z}\,dw,% \qquad f_{1}(z)=-\frac{1}{2\pi i}\int_{|z-z_{0}|=r^{\prime}}\frac{f(w)}{w-z}\,dw.

Then for $r^{\prime}<|z-z_{0}|<R^{\prime}$ we have $f(z)=f_{0}(z)+f_{1}(z)$ ; $f_{0}$ is analytic on $|z-z_{0}|<R^{\prime}$ , and $f_{1}$ is analytic on $|z-z_{0}|>r^{\prime}$ with the path viewed as the boundary of $\{z:|z-z_{0}|>r^{\prime}\}$ , and is analytic at infinity with value $0$ since $\lim_{z\to\infty}f_{1}(z)=0$ . (One could also make a more formal argument by evaluating on $1/z$ and taking a limit.) We are still using $r^{\prime}$ and $R^{\prime}$ instead of $r$ and $R$ , but could conclude by taking the limit as $r^{\prime}\to r$ and $R^{\prime}\to R$ , or by using the uniqueness part above to observe that the result is independent of $r^{\prime}$ and $R^{\prime}$ . ∎

Before proceeding let’s work out an example, which demonstrates the importance of fixing the annulus. Consider $f(z)=\frac{1}{(z-1)(z-2)}$ and $z_{0}=0$ . There are at least three different annuli we could consider on which $f$ is analytic: $0<|z|<1$ ; $1<|z|<2$ ; and $2<|z|$ .

On the first annulus, which is a punctured disk, note that $f$ actually extends to an analytic function on $|z|<1$ , so we can take $f_{0}(z)=f(z)$ and $f_{1}(z)=0$ .

Similarly on the third annulus, $f$ is analytic at infinity with value $\lim_{z\to\infty}f(z)=0$ , so we can just take $f_{0}(z)=0$ and $f_{1}(z)=f$ .

The most interesting case is the second annulus, where we cannot extend to $|z|<2$ since we have a pole at $z=1$ , nor to $|z|>1$ since we have a pole at $2$ . Instead, we can use the partial fraction decomposition:

f(z)=\frac{1}{(z-1)(z-2)}=\frac{A}{z-1}+\frac{B}{z-2}=\frac{A(z-2)+B(z-1)}{(z-% 1)(z-2)}=\frac{(A+B)z-(2A+B)}{(z-1)(z-2)}

implies $B=-A$ and so $A=-1$ , so

f(z)=-\frac{1}{z-1}+\frac{1}{z-2}.

We note that $f_{0}(z)=\frac{1}{z-2}$ is analytic on $|z|<2$ and $f_{1}(z)=-\frac{1}{z-1}$ is analytic on $|z|>1$ , and is analytic at infinity with value $0$ , so $f(z)=f_{0}(z)+f_{1}(z)$ has the desired properties.

In particular, note that the decomposition looks different depending which annulus we take.

Suppose $f$ is analytic on our annulus, so by our proposition we can write $f(z)=f_{0}(z)+f_{1}(z)$ with $f_{0}$ , $f_{1}$ satisfying the properties above. Since $f_{0}$ is analytic on $|z|<R$ , we can write its Taylor series

f_{0}(z)=\sum_{n=0}^{\infty}a_{n}z^{n}.

Since $f_{1}$ is analytic at infinity, with $f(\infty)=0$ , we can write its Taylor series at infinity as

f_{1}(z)=\sum_{n=1}^{\infty}b_{n}z^{-n}=\sum_{n=-\infty}^{-1}b_{-n}z^{n}.

Putting these together, if we write $a_{-n}=b_{n}$ for $n\geq 1$ , we get

f(z)=f_{0}(z)+f_{1}(z)=\sum_{n=-\infty}^{\infty}a_{n}z^{n}

for $z\in D$ , i.e. an expansion of $f$ as a linear combination of $z^{n}$ over all integers $n$ , as we claimed we’d find. This is the Laurent expansion of $f$ on $D$ with respect to $z_{0}$ .

We’d like to have a formula for the coefficients $a_{n}$ in terms of $f$ . For the Taylor series expansion—equivalently, the Laurent series in the special case where all $a_{n}=0$ for $n<0$ —we had

a_{n}=\frac{1}{n!}f^{(n)}(z_{0})=\frac{1}{2\pi i}\int_{|z-z_{0}|=R^{\prime}}% \frac{f(z)}{(z-z_{0})^{n+1}}\,dz

for some $R^{\prime}<R$ . The first formula doesn’t make sense when $n$ is negative, but the second formula actually still does, provided $r<R^{\prime}<R$ , and the same proof applies:

	$\displaystyle\frac{1}{2\pi i}\int_{\|z-z_{0}\|=R^{\prime}}\frac{f(z)}{(z-z_{0})^% {n+1}}\,dz$	$\displaystyle=\frac{1}{2\pi i}\int_{\|z-z_{0}\|=R^{\prime}}\sum_{m=-\infty}^{% \infty}a_{m}z^{m-n-1}\,dz$
		$\displaystyle=\frac{1}{2\pi i}\sum_{m=-\infty}^{\infty}a_{m}\int_{\|z-z_{0}\|=R^% {\prime}}z^{m-n-1}\,dz,$

and the inner integral is $0$ unless $m-n-1=-1$ , i.e. $m=n$ , in which case it is $2\pi i$ . Hence the whole sum is just $a_{n}$ as desired, whether $n$ is positive or negative.

In particular, since the coefficients are determined by the values of the function, the Laurent series expansion on a given annulus around a central point is unique (this would also follow from the uniqueness arguments above).

Let’s return to our previous example of $f(z)=\frac{1}{(z-1)(z-2)}$ on the most interesting annulus $1<|z|<2$ . Writing

f(z)=f_{0}(z)+f_{1}(z)=\frac{1}{z-2}-\frac{1}{z-1},

we have

f_{0}(z)=\frac{1}{z-2}=-\frac{1}{2}\cdot\frac{1}{1-z/2}=\sum_{n=0}^{\infty}-% \frac{1}{2^{n+1}}z^{n}

around $z=0$ , and

f_{1}(z)=-\frac{1}{z-1}=-\frac{1/z}{1-1/z}=\sum_{n=1}^{\infty}-z^{-n},

f(z)=\sum_{n=-\infty}^{\infty}a_{n}z^{n}

where $a_{n}=-\frac{1}{2^{n+1}}$ if $n\geq 0$ and $n=-1$ if $n<0$ .

We could also expand $f$ about one of the poles. Consider for example $z_{0}=1$ . Writing

f(z)=\frac{1}{(z-1)(z-2)}=\frac{1}{z-1}\cdot\frac{1}{(z-1)-1},

by the geometric series this is

-\frac{1}{z-1}\sum_{n=0}^{\infty}(z-1)^{n}=\sum_{n=-1}^{\infty}-(z-1)^{n}

for $0<|z-1|<1$ .