CVS1

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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\,.


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