Let \(f(z)=1/(z(1+z^2))\).
Find a Laurent series representing \(f\) for \(|z|<1\).
Find another Laurent series representing \(f\) for \(|z|>1\).
Solution
1. On\[ 0<|z|<1, \]one has\[ \frac1{1+z^2} = \sum_{n=0}^{\infty}(-1)^n z^{2n}. \]
For \(|z|<1\), one has \(|z^2|<1\), so the geometric-series identity \[ \frac1{1-w} = \sum_{n=0}^{\infty}w^n \] applied to \(w=-z^2\) gives \[ \frac1{1+z^2} = \sum_{n=0}^{\infty}(-z^2)^n = \sum_{n=0}^{\infty}(-1)^n z^{2n}. \] The puncture at \(z=0\) is required because \(f\) itself has a pole there.
2. Therefore, for\[ 0<|z|<1, \]the Laurent series of \(f\) is\[ \boxed{ f(z) = \sum_{n=0}^{\infty}(-1)^n z^{2n-1}. } \]
By step 1, \[ \begin{aligned} f(z) &= \frac1z\frac1{1+z^2}\\ &= \frac1z \left( \sum_{n=0}^{\infty}(-1)^n z^{2n} \right)\\ &= \sum_{n=0}^{\infty}(-1)^n z^{2n-1}. \end{aligned} \]
3. On\[ |z|>1, \]one has\[ \frac1{1+z^{-2}} = \sum_{n=0}^{\infty}(-1)^n z^{-2n}. \]
If \(|z|>1\), then \[ |z^{-2}|<1. \] Applying the geometric series to \(w=-z^{-2}\) gives the displayed identity.
4. Therefore, for\[ |z|>1, \]the Laurent series of \(f\) is\[ \boxed{ f(z) = \sum_{n=0}^{\infty}(-1)^n z^{-2n-3}. } \]
Rewrite \[ f(z) = \frac1{z(1+z^2)} = \frac1{z^3}\frac1{1+z^{-2}}. \] Using step 3, \[ \begin{aligned} f(z) &= \frac1{z^3} \left( \sum_{n=0}^{\infty}(-1)^n z^{-2n} \right)\\ &= \sum_{n=0}^{\infty}(-1)^n z^{-2n-3}. \end{aligned} \]
5. Q.E.D.
Steps 2 and 4 give the two requested Laurent expansions and their domains.