Laurent series complex analysis

Question

I'm meant to find the laurent series for this function;

$$f(z) = \frac{1}{z^2-1}, 1<|z-2|<3 $$

I started with substituting $$ t=z-2 \implies f(t+2) = \frac{1}{(t+1)(t+3)}$$

And then I did used partial fraction so I have 2 series to work with. My two series looked like this $$\frac{1}{2}\sum_{n=0}^\infty(-1)^n(z-2)^n - \frac{1}{2}\sum_{n=0}^\infty(-\frac{1}{3})^n(z-2)^n$$

Is it correct? When I add them together, I get a new sum which is the wrong answer. The correct answer also involves a sum from negative to positive infinity. I don't understand how, since both of the sums I got start from n=0.

Answer 1

Have a look! $$\begin{align}f(z) &= \frac{1}{z^2-1}\\ &=\frac{1}{(z-1)(z+1)}\\ &=\frac{1}{2(z-1)}-\frac{1}{2(z+1)}\\ &=\frac{1}{2[(z-2)+1]}-\frac{1}{2[(z-2)+3]}\\ &=\frac{1}{2(z-2)[1+1/(z-2)]}-\frac{1}{2.3[1+(z-2)/3]}\\ &=\frac{1}{2(z-2)}\sum _{n=0}^{\infty } \frac{(-1)^n}{(z-2)^n}-\frac{1}{2.3}\sum _{n=0}^{\infty } \frac{(-1)^n(z-2)^n}{3^n}\\ &=\frac{1}{2}\sum _{n=0}^{\infty } \frac{(-1)^{n}}{(z-2)^{n+1}}-\frac{1}{2}\sum _{n=0}^{\infty } \frac{(-1)^n(z-2)^n}{3^{n+1}}\\ &=\frac{1}{2}\sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{(z-2)^{n}}-\frac{1}{2}\sum _{n=0}^{\infty } \frac{(-1)^n(z-2)^n}{3^{n+1}}\\ &=\frac{1}{2}\sum _{n=-\infty}^{-1 } (-1)^{(1-n)}(z-2)^{n}-\frac{1}{2}\sum _{n=0}^{\infty } \frac{(-1)^n(z-2)^n}{3^{n+1}}\\ &=\frac{1}{2}\sum _{n=-\infty}^{\infty } \Bigg(**\Bigg){(z-2)^n} \end{align}$$