9.2 Examples of Laurent Series

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Given a specific function $f$ that is holomorphic on an annulus, we want to be calculate the Laurent series of $f$; that is, we want to be able to calculate the coefficients $a_n$. If we were to appeal directly to Laurent's theorem we would have to evaluate many contour integrals. Instead, we can appeal to the following lemma, giving uniqueness of coefficients.

Lemma

If two Laurent series $$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-b{)}^n\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n(z-b{)}^n$$ are equal on $\mathsf{Ann}(b,r,R)$ then $a_n=c_n$ for all $n\in \mathbb{Z}$.

Proof:

Define $$\begin{array}{rl}f(z)& =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-b{)}^n-\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n(z-b{)}^n\\ & =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(a_n-c_n)(z-b{)}^n\end{array}$$ on $\mathsf{Ann}(b,r,R)$. By hypothesis $f(z)=0$ for all $z\in \mathsf{Ann}(b,r,R)$. Then Laurent's theorem gives $$a_n-c_n={\displaystyle \frac{1}{2\pi i}}\underset{\gamma }{\int }{\displaystyle \frac{f(z)}{(z-b{)}^{n+1}}}\mathsf{d}z=0$$ for all $n\in \mathbb{Z}$. $◻$

We conclude this section with the calculation of some Laurent series.

Example

Let $f(z)=\mathsf{exp}(z)+\mathsf{exp}(1/z)$. Recall that $$\mathsf{exp}(z)=\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{z^n}{n!}}$$ for all $z\in \mathbb{C}$. Hence $$\mathsf{exp}(1/z)=\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{z^n}}{\displaystyle \frac{1}{n!}}$$ for all $z\ne 0$ and $$f(z)=\cdots +{\displaystyle \frac{1}{n!}}{\displaystyle \frac{1}{z^n}}+\cdots +{\displaystyle \frac{1}{2!}}{\displaystyle \frac{1}{z^2}}+\frac{1}{z}+2+z+{\displaystyle \frac{z^2}{2}}+\cdots +\frac{z^n}{n!}+\cdots$$ for all $z\ne 0$. This is a Laurent series on $\mathbb{C}\setminus \{0\}$ and it is equal to $f(z)$ for all non-zero $z$ so, by the Lemma, it is the Laurent series of $f$ on $\mathsf{Ann}(0,0,\mathrm{\infty })$.

Example

Let $$f(z)=\frac{1}{z}+\frac{1}{1-z}$$ and let us calculate the Laurent series at $b=0$. Now $1/z$ is already a Laurent series at $0$ (with $a_{-1}=1$ the only non-zero coefficient). This converges whenever $z\ne 0$. From the geometric series $${\displaystyle \frac{1}{1-z}}=\sum _{n=0}^{\mathrm{\infty }}{z}^n$$ and this power series converges whenever $|z|<1$. Hence $f$ has Laurent series $$f(z)=\frac{1}{z}+1+z+z^2+z^3+\cdots =\sum _{n=-1}^{\mathrm{\infty }}{z}^n$$ and this expression is valid on the annulus $\mathsf{Ann}(0,0,1)$.

Example

Let $$f(z)=\frac{1}{z-1}-\frac{1}{z-2}$$ on $\mathbb{C}\setminus \{1,2\}$. We will give three different Laurent series for $f$ valid in three different annuli.

First note that we can write $$\frac{1}{z-1}=\frac{-1}{1-z}=-\sum _{n=0}^{\mathrm{\infty }}{z}^n$$ whenever $|z|<1$ using geometric series. We can also write $$\frac{1}{z-1}=\frac{1}{z}\frac{1}{(1-\frac{1}{z})}=\frac{1}{z}\sum _{n=0}^{\mathrm{\infty }}{z}^{-n}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{z^n}$$ by another geometric series expansion, valid this time whenever $|1/z|<1$ i.e. whenever $|z|>1$.

Similarly, we can write $$\frac{1}{z-2}=\frac{-1}{2-z}=-\frac{1}{2}\left(\frac{1}{1-\frac{z}{2}}\right)=-\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{z}{2}\right)}^n$$ whenever $|z/2|<1$ i.e. whenever $|z|<2$, and we can write $$\frac{1}{z-2}=\frac{1}{z}\frac{1}{(1-\frac{2}{z})}=\frac{1}{z}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{2}{z}\right)}^n=\sum _{n=0}^{\mathrm{\infty }}\frac{2^n}{z^{n+1}}$$ whenever $|2/z|<1$ i.e. whenever $|z|>2$.

From the above we have $$f(z)=\{\begin{array}{ll}-{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{z}^n+\frac{1}{2}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{z}{2}\right)}^n={\displaystyle \sum _{n=0}^{\mathrm{\infty }}(-1+\frac{1}{2^{n+1}})z^n}}}& z\in \mathsf{Ann}(0,0,1)\\ {\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{z^n}}+\frac{1}{2}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{z}{2}\right)}^n}}& z\in \mathsf{Ann}(0,1,2)\\ {\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{1}{z^{n+1}}-{\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{2^n}{z^{n+1}}={\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{1-2^n}{z^{n+1}}}}}& z\in \mathsf{Ann}(0,2,\mathrm{\infty })\end{array}$$ giving the desired expansions.

In the above examples we have expanded functions as Laurent series on annuli centred at the origin. If we want to expand a function $f(z)$ as a Laurent series on an annulus centred at some other point $b$ then it is often convenient to first change co-ordinates to $w=z-b$, calculate the Laurent series in terms of $w$, and then change co-ordinates back to $z$.

Example

Let $$f(z)=\frac{\mathsf{exp}(z)}{(z-1{)}^2}.$$ We will expand $f$ as a Laurent series on the annulus $\mathsf{Ann}(1,0,\mathrm{\infty })$. We first change co-ordinates via $w=z-1$. Then $z=1+w$ and we are interested in expanding $$g(w)={\displaystyle \frac{\mathsf{exp}(1+w)}{w^2}}$$ on $\mathsf{Ann}(0,0,\mathrm{\infty })$. We have $${\displaystyle \frac{\mathsf{exp}(1+w)}{w^2}}={\displaystyle \frac{e}{w^2}}\sum _{n=0}^{\mathrm{\infty }}\frac{w^n}{n!}=\frac{e}{w^2}+\frac{e}{w}+\frac{e}{2!}+\frac{e}{3!}w+\frac{e}{4!}{w}^2+\cdots +\frac{e}{n!}{w}^{n-2}+\cdots$$ whenever $w\ne 0$. Going back to $z$ gives $$f(z)=g(z-1)=\frac{e}{(z-1{)}^2}+\frac{e}{z-1}+\frac{e}{2!}+\frac{e}{3!}(z-1)+\frac{e}{4!}(z-1{)}^2+\cdots +\frac{e}{n!}(z-1{)}^{n-2}+\cdots$$ for all $z\ne 1$ i.e. on $\mathsf{Ann}(1,0,\mathrm{\infty })$.