Find the Laurent expansions about \(z=0\) of the following functions: \[ e^{1\over z} \hspace{8em} \cos \qty{1\over z} .\]
Solution
Substitute \(1/z\) into the usual Taylor series.
For the exponential, \[ e^{1/z} =\sum_{n=0}^\infty {1\over n!z^n} =1+{1\over z}+{1\over2!z^2}+{1\over3!z^3}+\cdots. \]
For the cosine, \[ \cos\qty{1\over z} =\sum_{n=0}^\infty {(-1)^n\over(2n)!z^{2n}} =1-{1\over2!z^2}+{1\over4!z^4}-{1\over6!z^6}+\cdots. \]
Both Laurent series converge for every \(z\neq0\), so their annulus of convergence is \(0<|z|<\infty\).