Find a Laurent expansion for \(f(z) \da {1\over (z-3)(z-1)}\) on the 3 annular regions centered at \(0\) where \(f\) is holomorphic.
Solution
The three regions are
\(0 \leq \abs{z} < 1\)
\(1 < \abs{z} < 3\)
\(3 < \abs{z} < \infty\)
Write \(f\) in terms of its principal parts at \(z=1\) and \(z=3\) by computing the residues:
\(\Res_{z=1}f(z) = (z-1)f(z)\evalfrom_{z=1} = {1\over z-3}\evalfrom_{z=1} = -{1\over 2}\)
\(\Res_{z=3}f(z) = (z-3)f(z)\evalfrom_{z=3} = {1\over z-1}\evalfrom_{z=3} = {1\over 2}\)
Thus \[ f(z) = {-1/2 \over z-1} + {1/2 \over z-3} .\]
Now find the two expansions for each term:
\[ {-1/2 \over z-1} &= {1/2 \over 1-z} = {1\over 2}\sum_{k\geq 0} z^k && 0 < \abs{z} < 1 \\ {-1/2 \over z-1} &= -{1\over 2}{\inverseof{z} \over \inverseof{z} - 1} = -{1\over 2z}{1\over 1-\inverseof{z}} = -{1\over 2}\sum_{k\geq 0}z^{-k-1} && 1 < \abs{z} < \infty \\ {1/2\over z-3} &= -{1\over 2}{1\over 3-z} = -{1\over 6}{1\over 1-{z\over 3}} = -{1\over 6}\sum_{k\geq 0}3^{-k} z^k && 0 < \abs{z} < 3 \\ {1/2\over z-3} &= {1\over 2z}{1\over 1-3\inverseof{z}} = {1\over 2z} \sum_{k\geq 0}3^kz^{-k} = {1\over 2}\sum_{k\geq 0}3^k z^{-k-1} && 3 < \abs{z} < \infty .\]
On each region, combine the expansions of the two terms that converge there: \[ 0 \leq \abs{z} < 1 \qquad & f(z) = {1\over 2}\sum_{k\geq 0}z^k - {1\over 6}\sum_{k\geq 0} 3^{-k}z^k \\ 1 < \abs{z} < 3 \qquad & f(z) = -{1\over 2}\sum_{k\geq 0}z^{-k-1} - {1\over 6}\sum_{k\geq 0} 3^{-k}z^k \\ 3 < \abs{z} < \infty \qquad & f(z) = - {1\over 2}\sum_{k\geq 0}z^{-k-1} + {1\over 2}\sum_{k\geq 0} 3^{k}z^{-k-1} .\]