# How to determine the Laurent series for $\frac{1}{z-2}$ for $|z|<2$

How to determine the Laurent series for $$\frac{1}{z-2}$$ for $|z|<2$?

I can write $$\frac{1}{z-2}=\frac{-1}{2-z}=\frac{-1}{2}\frac{1}{1-\frac{z}{2}}=\frac{-1}{2} \sum_{n=0}^{\infty}\left(\frac{z}{2}\right)^n=-\sum_{n=0}^{\infty}\frac{z^n}{2^{n+1}}$$ for $|z|<2$, I think this first part is correct although I'm not exactly why this is a Laurent series because all the exponents are positive so shouldn't this be a Taylor series?

Also what about if I had $$\frac{1}{(z-2)(z-3)}$$ and I wanted to find a Laurent series on the same disc as before. I would use partial fractions to split it up and do the same with the $z-2$ part as before but how would I deal with the $z-3$ bit because that wouldn't have convergence in $|z|<2$ would it?

Indeed, the series is also a Taylor series. What happens is that all Taylor series are Laurent series but the converse doesn't hold. You can simply write $$-\sum_{n=0}^{\infty}\frac{z^n}{2^{n+1}}=\sum_{n=-\infty}^\infty c_nz^n,$$ taking $c_n=-1/2^{n+1}$ for $n\ge0$ and $c_n=0$ for $n<0$, and you get a Laurent series on the right-hand side.
For the other function, note that $$\frac{1}{(z-2)(z-3)}=\frac{1}{z-3}-\frac{1}{z-2}$$ and $$\frac{1}{z-3}=\frac{-1/3}{1-z/3}.$$ Since $|z/3|\le 2/3<1$ for $|z|<2$, you can do the same as before.
• Thanks. I understand better your answer was very good. One last question what about if instead of $z-3$ we had $z-1$ then we would get the $z-1$ part would converge for $|z|<1$ as opposed to $|z/3|<1$ so is it even possible to find a Laurent series if we replaced $z-3$ with $z-1$? – Damien Mar 3 '16 at 21:23
• More or less the same, you need to separate into two cases. For $|z|<1$ you can write $1/(z-1)=-1/(1-z)$ and for $1<|z|<2$ you can write$$\frac1{z-1}=\frac{1/z}{1-1/z}$$ and note that $|1/z|<1$. – John B Mar 3 '16 at 21:26