Proving a function is analytic

Asked Jan 31 '17 at 04:52

Active Jan 31 '17 at 04:52

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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

asked Jan 31 '17 at 04:52

paranoidhominid

What is expansion of $e^{\frac1z}$.? – Nosrati Jan 31 '17 at 04:54
@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? – paranoidhominid Jan 31 '17 at 05:04
Are you supposed to show directly that $e^{1/z}$ is harmonic for $z\ne 0$? – Mark Viola Jan 31 '17 at 05:06
See e.g. Composition of a harmonic function with a holomorphic function for a general result – Winther Jan 31 '17 at 05:09
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$\exp$ is an entire function, and $1/z$ is analytic except at $0$, so... – Robert Israel Jan 31 '17 at 05:10
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. – juan arroyo Jan 31 '17 at 17:17

Proving a function is analytic

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