How do I find the Laurent series expansion for this complex function?

Question

I have given $f(z)=\frac{1}{z^2-1}$ and I want to find the Laurent series expansion for $0<|z-1|<2$.

I have just remarked that $$f(z)=\frac{\frac{1}{2}}{z-1}-\frac{\frac{1}{2}}{z+1}$$I know that I somehow need to rewrite the fractions in order that I can use the well known geometric series, but I do not see how I can do this using that $0<|z-1|<2$.

Could maybe someone help me?

Answer 1

The term $\dfrac{\frac{1}{2}}{z-1}$ is already a term, of degree $-1$, of the desired Laurent series. Then for the other fraction, as you said, use a geometric series expansion (I'm dropping the $\frac{1}{2}$ from the numerator, so you should multiply back by it): $$\frac{1}{z+1}=\frac{1}{2+(z-1)}=\frac{1}{2}\cdot\frac{1}{1+\frac{z-1}{2}}=\frac{1}{2}\cdot\frac{1}{1-\left(-\frac{z-1}{2}\right)}=\frac{1}{2}\cdot\sum_{n=0}^{\infty}\left(-\frac{z-1}{2}\right)^n=\cdots,$$ and then simplify this a bit further.

Note that this geometric series converges for $\left|-\dfrac{z-1}{2}\right|<1$, i.e. for $|z-1|<2$, as desired.

Answer 2

Since\begin{align}\frac1{z+1}&=\frac1{2+(z-1)}\\&=\frac12\cdot\frac1{1+\frac{z-1}2}\\&=\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}(z-1)^n,\end{align}you have\begin{align}\frac1{z^2-1}&=(z-1)^{-1}\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}(z-1)^n\\&=\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}(z-1)^{n-1}\\&=\sum_{n=-1}^\infty\frac{(-1)^{n+1}}{2^{n+2}}(z-1)^n.\end{align}