Find the Laurent series expansions centered at \(z_0 = 0\) for the rational function: \[f(z) = \frac{1}{1 - z^2} + \frac{1}{3 - z}.\] (1) How many distinct Laurent series expansions centered at \(0\) exist for \(f(z)\)? (2) State the annular domain of convergence for each expansion and write the explicit series.
Solution
The poles are at \(z=\pm1\) and \(z=3\). Their distances from the center \(0\) are \(1\) and \(3\), so there are three maximal annuli of holomorphy centered at \(0\): \[ |z|<1, \qquad 1<|z|<3, \qquad |z|>3. \] Thus there are exactly three Laurent expansions centered at \(0\).
1. For \(|z|<1\),\[ \frac1{1-z^2}=\sum_{n=0}^\infty z^{2n}, \qquad \frac1{3-z}=\sum_{n=0}^\infty\frac{z^n}{3^{n+1}}. \]Hence\[ f(z)=\sum_{n=0}^\infty z^{2n} +\sum_{n=0}^\infty\frac{z^n}{3^{n+1}}. \]
2. For \(1<|z|<3\),\[ \frac1{1-z^2} =-\sum_{n=1}^\infty z^{-2n}, \qquad \frac1{3-z}=\sum_{n=0}^\infty\frac{z^n}{3^{n+1}}, \]so\[ f(z)=-\sum_{n=1}^\infty z^{-2n} +\sum_{n=0}^\infty\frac{z^n}{3^{n+1}}. \]
3. For \(|z|>3\),\[ \frac1{1-z^2}=-\sum_{n=1}^\infty z^{-2n}, \qquad \frac1{3-z}=-\sum_{n=1}^\infty3^{n-1}z^{-n}. \]Therefore\[ f(z)=-\sum_{n=1}^\infty z^{-2n} -\sum_{n=1}^\infty3^{n-1}z^{-n}. \]