Find the Laurent expansion for $$\frac{\exp(\frac 1{z^2})}{z-1}$$ about $z=0$. I know that $\exp( \frac 1{z^2}) = \sum_{n=0}^\infty \frac{z^{-2n}}{n!}$ and $ \frac 1{z-1}=-\sum_{n=0}^\infty z^n$
Laurent Series of rational function about $z=0$
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complex-analysis
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1And do you know how to multiply power series: http://en.wikipedia.org/wiki/Power_series#Multiplication_and_division ? – 2012-03-20
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0I was thinking of that but I was having difficult time simplifying it and finding a direct formula for the coefficients. Does it matter that the power series we are multiplying have different powers of z because in the given example they both have (x-c)^n – 2012-03-20