laurent series of 1/(z(z-1)(z+1) in |z|>1

Question

I want to compute the laurent series expansion of $\frac{1}{z(z-1)(z+1)}$ in the annulus $|z|>1$. I'm a little confused at having expansions at both $\pm1$ - can I still handle it as usual, or do I need to do anything to account for this?

Thanks!

Answer 1

Hint:

$$\frac1{2z}\left(\frac1{z-1}-\frac1{z+1}\right)=\frac1{2z}\left(\frac1z\cdot\frac1{1-\frac1z}-\frac1z\cdot\frac1{1+\frac1z}\right)=$$

$$=\frac1{2z^2}\left(\frac1{1-\frac1z}-\frac1{1+\frac1z}\right)$$

Now use geometric series and the fact that $\;\left|\cfrac1z\right|<1\;$