Let \(\displaystyle f(z) = \frac{1}{z} + \frac{1}{z^2 -1}\). Find all the Laurent series of \(f\) and describe the largest annuli in which these series are valid.
Solution
The singularities are at \(-1,0,1\), so the maximal annuli centered at \(0\) are \[ 0<|z|<1 \qquad\text{and}\qquad |z|>1. \]
For \(0<|z|<1\), \[ \frac1{z^2-1}=-\frac1{1-z^2}=-\sum_{n=0}^\infty z^{2n}, \] and hence \[ \boxed{ f(z)=\frac1z-\sum_{n=0}^\infty z^{2n} =\frac1z-1-z^2-z^4-\cdots, \qquad 0<|z|<1.} \]
For \(|z|>1\), \[ \frac1{z^2-1} =\frac1{z^2}\frac1{1-z^{-2}} =\sum_{n=0}^\infty z^{-2n-2}, \] so \[ \boxed{ f(z)=\frac1z+\sum_{n=0}^\infty z^{-2n-2} =\frac1z+\frac1{z^2}+\frac1{z^4}+\frac1{z^6}+\cdots, \qquad |z|>1.} \] These are all Laurent expansions about \(0\) on maximal annuli of analyticity.