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Show that $f(z)=e^{1/z}$ is analytic, except when $z = 0$

Since $$\Delta f = \bigtriangledown ^2f = \frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} $$ my guess is that I should try to express $f(z)$ in terms of $x + iy$? Thanks for your time

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    What is expansion of $e^{\frac1z}$.? – 2017-01-31
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    @MyGlasses $$e^{\frac{1}{z}} = \sum_{n=0}^\infty \frac{1}{z^nn!}$$ and from what I looked up, that must be equal to its Taylor series, right? – 2017-01-31
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    Are you supposed to show directly that $e^{1/z}$ is harmonic for $z\ne 0$? – 2017-01-31
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    See e.g. [Composition of a harmonic function with a holomorphic function](http://math.stackexchange.com/questions/151827/composition-of-a-harmonic-function-with-a-holomorphic-function) for a general result – 2017-01-31
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    $\exp$ is an entire function, and $1/z$ is analytic except at $0$, so... – 2017-01-31
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    You have to prove that a function is holomorphic or analytic first before you can conclude that it equals its Taylor series. That's why the definition "analytic" exists, because not all functions equal their Taylor series. – 2017-01-31

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