Find the Laurent expansion about \(z=0\) and \(z=1\) respectively of the following function: \[ f(z) \da {z+1 \over z(z-1)} .\]
Solution
Note: once you see that everything is in terms of powers of \((z-z_0)\), you’re essentially done. For \(z=0\): \[ {z+1 \over z(z-1)} &= {1\over z} {z+1 \over z-1} \\ &= -{z+1\over z} {1\over 1-z} \\ &= -\qty{1 + {1\over z}}\sum_{k\geq 0} z^k .\]
For \(z=1\): \[ {z+1 \over z(z-1)} &= {1\over z-1}\qty{1 + {1\over z} } \\ &= {1\over z-1}\qty{1 + {1\over 1 - (1-z)} } \\ &= {1\over z-1} \qty{1 + \sum_{k\geq 0} (1-z)^k } \\ &= {1\over z-1} \qty{1 + \sum_{k\geq 0} (-1)^k (z-1)^k } .\]