CVS1

Find the Laurent series for $f(z) = z^2 e^{1/z}\,$ about the singular point $z=0\,$ .

Use the series $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}, |z|<\infty\,$ .

$z^2 e^{1/z} = z^2 \sum_{n=0}^\infty\frac{(1/z)^n}{n!}\,$

$=z^2+z+\frac{1}{2!} + \frac{1}{3!z} + \frac{1}{4!z^2} + ..., |z|>0\,$

The domain of convergence is all $|z|>0\,$ .