Expand \(\frac{1}{1-z^{2}}+\frac{1}{z-3}\) in a series of the form \(\sum_{-\infty}^{\infty} a_{n} z^{n}\) so it converges for
\(|z|<1\),
\(1<|z|<3\),
\(|z|>3\).
Solution
General strategy: each has two expansions, so just compute them all and pick appropriate ones for regions afterwards.
For \(1\over z-3\): \[ {1\over z-3} &= -{1\over 3}{1\over 1- {z\over 3}} = -{1\over 3}\sum_{k\geq 0}3^{-k}z^k && \abs{z} < 3 \\ &= {1\over z} {1\over 1 - {3\over z}} = \inverseof{z} \sum_{k\geq 0} 3^k z^{-k} && \abs{z} > 3 .\]
For \(1\over 1-z^2\): \[ {1\over 1-z^2} &= \sum_{k\geq 0} z^{2k} && \abs{z} < 1 \\ &= {1\over z^2} {-1\over 1- z^{-2}} = -z^{-2}\sum_{k\geq 0}z^{-2k} && \abs{z} > 1 .\]
So take \[ 0 < \abs{z} < 1 && f(z) &= \sum_{k\geq 0}z^{2k} - {1\over 3}\sum_{k\geq 0} 3^{-k}z^k \\ 1 < \abs{z} < 3 && f(z) &= -z^{-2} \sum_{k\geq 0}z^{-2k} - {1\over 3}\sum_{k\geq 0} 3^{-k}z^k \\ 3 < \abs{z} < \infty && f(z) &= -z^{-2} \sum_{k\geq 0}z^{-2k} + \inverseof{z} \sum_{k\geq 0}3^k z^{-k} .\]