# Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$

We have $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$. I want to expand this as a Laurent series in $z_0=2$ on $\{4<|z-2|<\infty\}$.

The partial decomposition is:

$$f(z)=\frac{1}{4}\frac{1}{z-2}-\frac{1}{4}\frac{1}{z+2}+\frac{1}{6-z}$$

In my reference, they expand $\frac{1}{z+2}$ and $\frac{1}{6-z}$ with the help of the geometric series, but they leave $\frac{1}{z-2}$ as it is. Why?

-
Beg pardon, but what's the reference you're using? – Alex Nelson Jul 24 '12 at 0:19
Some exercise notes from a complex analysis course I attended. – Chris Jul 24 '12 at 0:21
"Expand this as a Laurent series" isn't enough data to know what to do on its own; you also need to know where the expansion is valid. If they're leaving $\frac{1}{z-2}$ alone, that's probably because they want an expansion which is valid on some annulus centered around $z=2$ (so the other things should all be expanded as powers of $z-2$). – Micah Jul 24 '12 at 0:24
I've added the missing information. So we don't do anything with $\frac{1}{z-2}$, because it's already part of the expansion around z=2 and there is nothing additional to be done here, right? – Chris Jul 24 '12 at 0:27
Yeah, you're expanding in terms of powers of $z-2$ and $\frac{1}{z-2}$ is already a power of $z-2$. (Also, I assume that by $\{|z-2|<\infty\}$ you mean $\{4<|z-2|<\infty\}$, which is the largest unbounded domain centered at $2$ on which $f$ is holomorphic?) – Micah Jul 24 '12 at 0:54

$$\frac{1}{z+2}=\frac{1}{(z-2)+4}=\frac{1}{4}\frac{1}{1+\left(\frac{z-2}{4}\right)}=\frac{1}{4}\left(1-\frac{z-2}{4}+\frac{(z-2)^2}{16}-\frac{(z-2)^3}{64}+\cdots\right)$$

$$\frac{1}{6-z}=-\frac{1}{(z-2)-4}=\cdots$$ and now just mimic the first development above of the given fraction in powers of $\,z-2\,$

Pay attention to the fact that the first (and also the second one, btw) development is valid for $$\left|\frac{z-2}{4}\right|<1\Longleftrightarrow|z-2|<4$$ Once you've done all the above just add the different fractions which form part of $\,f(z)\,$

-
Thank you very much! – Chris Jul 24 '12 at 2:29

To expand as a Laurent series about $z_0=2$, you want to write the function as a power series in $z-2$, possibly with negative powers of $z-2$. Therefore, we can use $w=z-2$ and $z=w+2$: \begin{align} \frac1{z^2-4}+\frac1{6-z} &=\frac{1}{w^2+4w}+\frac1{4-w}\\ &=\frac1w\left(\frac14-\frac1{16}w+\frac1{64}w^2-\frac1{256}w^3+\frac1{1024}w^4-\dots\right)\\ &+\left(\frac14+\frac1{16}w+\frac1{64}w^2+\frac1{256}w^3+\dots\right)\\ &=\frac1{4w}+\frac3{16}+\frac5{64}w+\frac3{256}w^2+\frac5{1024}w^3+\dots\\ &=\frac1{4(z-2)}+\frac3{16}+\frac5{64}(z-2)+\frac3{256}(z-2)^2+\frac5{1024}(z-2)^3+\dots \end{align} We have expanded the series in $w$ (aka $z-2$) about $w=0$ (aka $z=2$). Both of the power series in $w$ converge for $0<|w|<4$, which is $0<|z-2|<4$, which is $-2<z<2$ or $2<z<6$, which is $-2<z<6$ but $z\not=2$.

-

Here is a related problem. You are asked to find the Laurent series at $x=2$ in the region $|z-2|>4$. You have

$$f(z) = \frac{1}{(z-2)(z+2)} - \frac{1}{z-6} = \frac{1}{(z-2)((z-2)+4)} - \frac{1}{{(z-2)-4}}$$

$$= \frac{1}{(z-2)^2(1+\frac{4}{(z-2)})} - \frac{1}{(z-2)(1-\frac{4}{z-2})}$$

$$= \sum_{k=0}^{\infty} \frac{(-1)^k 4^k}{(z-2)^{k+2}}- \sum_{k=0}^{\infty} \frac{4^k}{(z-2)^{k+1}}$$

The above two series have the region of convergence $|z-2|>4\,.$ It is very important to pay attention to the region (in your case $|z-2|>4$) where you are asked to construct Laurent series and based on it you find the Laurent series.

-