Find the Laurent expansions of \(\exp ( \textstyle { \frac { 1 } { z } } )\) and \(\cos { \left( { \frac { 1 } { z } } \right) }\) about the origin.
Solution
1. The Laurent expansion of \(e^{1/z}\) about the origin is\[ \boxed{ e^{1/z} = \sum_{n=0}^{\infty}\frac{z^{-n}}{n!}. } \]It converges on the maximal annulus\[ 0<\abs{z}<\infty. \]
The exponential Taylor series \[ e^w = \sum_{n=0}^{\infty}\frac{w^n}{n!} \] converges for every \(w\in\CC\). For \(z\neq0\), substitute \[ w=\frac1z. \] This gives \[ e^{1/z} = \sum_{n=0}^{\infty}\frac{z^{-n}}{n!}. \] Since the Taylor series converges for every finite value of \(w=1/z\), the Laurent series converges for every \(z\neq0\). Thus its annulus of convergence about the origin is \(0<\abs{z}<\infty\).
2. The Laurent expansion of \(\cos(1/z)\) about the origin is\[ \boxed{ \cos(1/z) = \sum_{n=0}^{\infty} \frac{(-1)^n z^{-2n}}{(2n)!}. } \]It also converges on\[ 0<\abs{z}<\infty. \]
The cosine Taylor series \[ \cos w = \sum_{n=0}^{\infty} \frac{(-1)^n w^{2n}}{(2n)!} \] converges for every \(w\in\CC\). Substituting \(w=1/z\) for \(z\neq0\) gives \[ \cos(1/z) = \sum_{n=0}^{\infty} \frac{(-1)^n z^{-2n}}{(2n)!}. \] Again, global convergence in the variable \(w\) shows that this Laurent series converges for every \(z\neq0\), hence on the maximal annulus \(0<\abs{z}<\infty\).
3. Q.E.D.
Steps 1 and 2 give both requested Laurent expansions and their maximal annuli of convergence.