i am trying to find the laurent series expansion for
$$f(z)={z+2\over z-1}$$
on both $0<|z|<1$ and $|z|>1$. For $0<|z|<1$, my thought has been to rewrite ${z+2\over z-1}$ as ${3\over z-1}+1$, then note
$${3\over z-1}={-3\over 1-z}= -3(1+z+z^2+\ldots).$$
Adding $1$ I get the series $-2-3z-3z^2-\ldots -3z^n$. The wolfram solution suggests there should be some negative powers of $z$. Any help as to where I have gone wrong would be great!
So, your approach to $|z|<1$ is correct. As for $|z|>1$ notice that $\frac{1}{|z|}<1$.Then
$\frac{z+2}{z-1} = \frac{z}{z-1} + \frac{2}{z-1} = \frac{1}{1-1/z}+ \frac{1}{z}\frac{2}{1-1/z} = \sum_{n=0}^{\infty}{\frac{1}{z^{n}}} + \frac{2}{z}\sum_{n=0}^{\infty}{\frac{1}{z^{n}}} = 2\sum_{n=0}^{\infty}{\frac{1}{z^{n+1}}} + \sum_{n=0}^{\infty}{\frac{1}{z^{n}}} $
And then it follows the Laurent series.