Find the Laurent expansion of \[ f(z) = {z+1 \over z(z-1)^2} \] about \(z=0\) and \(z=1\) respectively.
Hint: recall that power series can be differentiated.
Solution
First decompose \[ \frac{z+1}{z(z-1)^2} =\frac1z-\frac1{z-1}+\frac2{(z-1)^2}. \]
About \(z=0\), for \(0<|z|<1\) we use \[ \frac1{1-z}=\sum_{n=0}^\infty z^n, \qquad \frac1{(1-z)^2}=\sum_{n=0}^\infty(n+1)z^n, \] to obtain \[ \boxed{ f(z)=\frac1z+\sum_{n=0}^\infty(2n+3)z^n, \qquad 0<|z|<1.} \] For \(|z|>1\) instead expand in powers of \(z^{-1}\): \[ \boxed{ f(z)=\sum_{n=2}^\infty(2n-3)z^{-n}, \qquad |z|>1.} \]
About \(z=1\), put \(\zeta=z-1\). Then \[ f(z)=\frac1{1+\zeta}-\frac1\zeta+\frac2{\zeta^2}. \] Thus, for \(0<|\zeta|<1\), \[ \boxed{ f(z)=\frac2{\zeta^2}-\frac1\zeta +\sum_{n=0}^\infty(-1)^n\zeta^n.} \] For \(|\zeta|>1\), \[ \boxed{ f(z)=\frac1{\zeta^2} +\sum_{n=3}^\infty(-1)^{n+1}\zeta^{-n}.} \] These are the Laurent expansions on the maximal annuli centered at the two specified points.