Give the Laurent series for the following functions, and characterize their singularities at zero (essential, removable, or pole):
\(\dfrac{\sin^2 z}{z}\)
\(ze^{-1/z^2}\)
\(\dfrac{1}{z(4-z)}\)
Solution
(a) Using \(\sin^2z=(1-\cos 2z)/2\), \[ \frac{\sin^2z}{z} =\sum_{k=1}^{\infty} (-1)^{k+1}\frac{2^{2k-1}}{(2k)!}z^{2k-1} =z-\frac{z^3}{3}+\frac{2z^5}{45}-\cdots. \] There are no negative powers, so the singularity at \(0\) is removable; the holomorphic extension takes the value \(0\) there.
(b) \[ ze^{-1/z^2} =z\sum_{k=0}^{\infty}\frac{(-1)^k}{k!z^{2k}} =\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}z^{1-2k} =z-\frac1z+\frac1{2z^3}-\frac1{6z^5}+\cdots. \] The Laurent series has infinitely many negative-power terms, so \(0\) is an essential singularity.
(c) For \(0<|z|<4\), \[ \frac1{z(4-z)} =\frac1{4z}\frac1{1-z/4} =\sum_{n=0}^{\infty}\frac{z^{n-1}}{4^{n+1}} =\frac1{4z}+\frac1{16}+\frac{z}{64}+\cdots. \] Thus \(0\) is a simple pole.