I believe $e^{z - \frac{1}{z}}$ has essential singularities at $z = 0$ and $z = \infty$ (in both cases because of a $\frac{1}{z}$ in the exponent) but I'm having a hard time proving this. How can one show this?
Singularities of $e^{z - \frac{1}{z}}$
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complex-analysis
power-series
singularity-theory
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0@JulienGodawatta the definition for singularities at infinity I'm using is the one given on http://en.wikipedia.org/wiki/Pole_(complex_analysis) – 2012-10-02
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0Forgive me, I've never seen this before. I'll delete my comment. – 2012-10-02
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0No problem, any ideas on how to go about this? – 2012-10-02
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0@Cocopuffs That's what I'd like to do but it seems ugly. I wrote $\sum \frac{1}{k!}(z-1/z)^k$ then expanded out the $(z-1/z)^k$ term using the binomial theorem, but that's a big mess. – 2012-10-02
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0Related: http://math.stackexchange.com/questions/233492/what-type-of-singularity-does-exp-fract2-z-frac1z-have-on-z – 2012-11-09