$e^\frac1z$ is not holomorphic at $z=0$, but it is known that it can be expanded as $$e^\frac1z=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots$$ The coefficients of this Laurent expansion are computed the same way as Taylor's. The question is how is that possible? If function is not holomoprhic at $z=0$, then it's not true that it is holomophic at $|z|
$e^{1/z}$ and Laurent expansion
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complex-analysis
laurent-series
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7Use the power series for $e^w$ and set $w=1/z$, knowing that it is invalid for $z=0$. Then use the uniqueness of the Laurent series. – 2012-11-06
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1Taylor series is always an analytic function throughout its disk of convergence. Laurent series is a generalization of Taylor series for functions with singularities. It just turn out that Laurent series looks like Taylor series in this case. – 2012-11-06
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3They are not Taylor coefficients. If $f(z)=e^{1/z}$, then $f^{(k)}(0)$ does not exist for $k\geq 0$. What is it that doesn't seem possible? There is a general integral formula for [Laurent series](http://en.wikipedia.org/wiki/Laurent_series) coefficients, but they are instead often found using some other known series, like in this case, where $f(z) = a_0 + a_1z +a_2z^2+\cdots$ is valid for all $z\in\mathbb C$, and it follows that $f(1/z)= a_0 + a_1/z +a_2/z^2+\cdots$ is valid for $z\neq 0$. Your $a_0$ should be $1$. – 2012-11-06
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0Can someone point me somewhere that derives the exponential series without using taylor series formulae? – 2014-11-18