Find the Laurent expansion of $\frac{z}{z^{2}-1}$ valid for $0<z-1<2$

Question

I have already used partial fractions and let $w=z-1$ to get to

$$\frac{1}{h}-\sum_{n=-1}^{\infty}{(-1)^{n+1}\frac{h^{n}}{2^{n+2}}}.$$

I know I can't have the $\frac{1}{h}$ in front of the sum. How do I get rid of this?

Answer 1

A Laurent series can start with a negative power (conventionally, a Taylor series should have only non-negative powers).

I would expand in $z-1$ because that seems to be what the question asks for when it says $0<z-1<2$ rather than $1<z<3$.

So the series becomes $$ \sum_{k=-1}^\infty a_k (z-1)^k $$ with $$ a_k =\left\{ \matrix{\frac12 & k= -1\\ \frac{(-1)^k}{2^{k+2}}& k\geq 0}\right. $$ Note that you must have made some mistake in your partial fraction decomposition (probably forgot that $z+1$ at the expansion point is $2$, not $1$), because the leading term is half what you were saying.

Answer 2

One way, perhaps clearer, to see it - Observe that $\;0<z-1<2\implies0<\frac{z-1}2<1\;$ , so using the geometric series development:

$$\frac z{(z-1)(z+1)}=\frac12\left(\frac1{z-1}+\frac1{z+1}\right)=\frac1{2(z-1)}+\frac12\frac1{z-1+2}=$$$${}$$

$$=\frac1{2(z-1)}+\frac14\frac1{1+\frac{z-1}2}=\frac1{2(z-1)}+\frac14\sum_{n=0}^\infty\frac{(-1)^n(z-1)^n}{2^n}$$

which, of course, is the same Mark has in his answer.