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$e^\frac1z$ is not holomorphic at $z=0$, but it is known it can be expanded as $$e^\frac1z=\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots$$ 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 $|z|<R$ and Taylor's coefficients can not be used. Please someone explain.

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Use 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. –  wj32 Nov 6 '12 at 2:47
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Taylor 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. –  glebovg Nov 6 '12 at 3:14
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They 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 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 $1$. –  Jonas Meyer Nov 6 '12 at 4:26

2 Answers 2

In general for $f$ entire, $$f(z)=a_0 +a_1z+a_2z^2+\ldots$$

From Cauchy's integral formula we have that $a_k=\frac{1}{2 \pi i} \oint_\gamma\frac{f(z)}{z^{k+1}}dz, \ k=\in \mathbb{Z}$ where $\gamma$ is the unit circle centered at zero($a_k=0$ for $k<0$).

The function $g(z)=f(\frac{1}{z})$ is analytic in $\mathbb{C}-\{0\}$ with Laurent series about zero $$g(z)=\ldots + \frac{b_{-2}}{z^2}+\frac{b_{-1}}{z}+b_0 +b_1z+b_2z^2+\ldots$$ with $b_k=\frac{1}{2 \pi i} \oint_\gamma\frac{g(z)}{z^{k+1}}dz, \ k=\in \mathbb{Z}$ where $\gamma$ is the unit circle centered at zero.

With substitution $z=\frac{1}{u}$ we get $b_k=\frac{1}{2 \pi i} \oint_\gamma\frac{g(\frac{1}{u})}{u^{-k-1}u^2}du=\frac{1}{2 \pi i} \oint_\gamma\frac{f(u)}{u^{-k+1}}du=a_{-k}.$

Therefore

$$g(z)=\ldots + \frac{a_{3}}{z^3} + \frac{a_2}{z^2}+\frac{a_1}{z}+a_0.$$

Or as wj32 said since $f(z)=a_0 +a_1z+a_2z^2+\ldots, \ \forall z \in \mathbb{C}$ we have $g(z)=f(\frac{1}{z})=\ldots + \frac{a_{3}}{z^3} + \frac{a_2}{z^2}+\frac{a_1}{z}+a_0 , \ \forall z \in \mathbb{C}- \{0\}$. From uniqueness of Laurent series this is the Laurent expansion of $g$.

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I'm pretty sure most of former answers didn't get the key point.

The Laurent series expansion is defined on a "deleted neighborhood" around a singularity, in this case, {z| 0<|z-0|< R }. In this deleted neighborhood, e^{1/z} is analytic. So for any point in this neighbourhood, we can expand e^{z} first and then substitue 1/z in.

As you can see in the Laurent expansion you gave, you can plug in any z arbitrarily close to zero, calculate the infinite sum, and get a finite and well defined value. The laurent series doesn't give any information about the behavior of a function "on" its singularities.

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