For the following functions, find the Laurent series about \(0\) and classify their singularities there: \[ {\sin^2(z) \over z} \\ z \exp{1\over z^2} \\ {1 \over z(4-z)} .\]
Solution
For the first function, use \(\sin^2z=(1-\cos 2z)/2\): \[ \frac{\sin^2z}{z} =\sum_{n=1}^\infty (-1)^{n+1}\frac{2^{2n-1}}{(2n)!}z^{2n-1} =z-\frac{z^3}{3}+\frac{2z^5}{45}-\cdots. \] There are no negative powers, so the singularity at \(0\) is removable; the holomorphic extension has value \(0\) there.
For the second function, \[ z e^{1/z^2} =z\sum_{n=0}^\infty\frac{1}{n!z^{2n}} =\sum_{n=0}^\infty\frac{z^{1-2n}}{n!} =z+\frac1z+\frac{1}{2!z^3}+\frac{1}{3!z^5}+\cdots. \] Its principal part has infinitely many nonzero terms, so \(0\) is an essential singularity.
Finally, for \(0<|z|<4\), \[ \frac1{z(4-z)} =\frac1{4z}\frac1{1-z/4} =\sum_{n=0}^\infty\frac{z^{n-1}}{4^{n+1}} =\frac1{4z}+\frac1{16}+\frac z{64}+\frac{z^2}{256}+\cdots. \] Thus \(0\) is a simple pole, with residue \(1/4\).