Find the Laurent expansion of \[ f(z) = {z + 1 \over z(z-1)} \] about \(z=0\) and \(z=1\) respectively.
Solution
Let \(f(z) = {z+1\over z(z-1)}\).
About \(z=0\):
\[ f(z) &= (z+1) \qty{- {1 \over z} + {1\over z-1} } \\ &= -(z+1) \qty{{1\over z} + \sum_{n=0}^\infty z^n } \\ &= -(z+1)\sum_{n=-1}^\infty z^n \\ &= {1\over z} + 2\sum_{n=0}^\infty z^n \\ &= -{1\over z} -2 - 2z - 2z^2 - \cdots .\]
About \(z=1\):
\[ f(z) &= \qty{(1-z) -2 \over 1-z} \qty{1 \over 1 - (1-z)} \\ &= \qty{1 - {2\over 1-z}} \sum_{n=0}^\infty (1-z)^n \\ &= \sum_{n=0}^\infty (1-z)^n - 2 \sum_{n=-1}^\infty (1-z)^n \\ &= -{2\over 1-z} - \sum_{n=0}^\infty (1-z)^n \\ &= {2\over z-1} + \sum_{n=0}^\infty (-1)^{n+1} (z-1)^n \\ &= {2\over z-1} - 1 + (z-1) - (z-1)^2 + \cdots .\]