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This is an exercise from Gamelin.

If $f(z)$ is a complex function with a not removable singularity in $ z_{0} \ $, then $e^{f(z)} \ $ has an essential singularity in $z_{0} $.

Any hint?

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$f(z)$ is complex, I guess... –  draks ... Apr 2 '12 at 9:27
    
@draks: yes, f is complex –  WLOG Apr 2 '12 at 9:29
    

2 Answers 2

up vote 4 down vote accepted

If your function $f$ has a pole at $z_0$ write $(z-z_0)^m f(z) = p(z) + (z-z_0)^m h(z)$ where $h$ is holomorphic and $p$ is a polynomial of degree $<m$. Then $f(z)= \displaystyle \frac{p(z)}{(z-z_0)^m} + h(z)$. Now take $e$ to both sides, note that $e^{h(z)}$ is holomorphic, and use the power series expansion of the exponential to show that $e^{\frac{p(z)}{(z-z_0)^m}}$ has infinitely many negative powers of $z-z_0$.

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Ok if f has a pole at z0 your solution is correct, but if f has an essential singularity in z0 ? –  WLOG Apr 2 '12 at 18:05
    
@stef That's the easy case. :) –  Parsa Apr 3 '12 at 17:06
    
How to show that this expansion necessarily has an infinitude of summands and that it cannot have some sort of telescoping series canceling the terms out...? –  Pranasas Nov 19 at 22:44

I haven't looked at this stuff in a while, but Casorati-Weierstrass might help.

In particular there is a sequence $(z_k)$ converging to $z_0$ such that $(f(z_k))$ converges to $w_0$ (this is page 175 of your book). Since $w_0$ is arbitrary can you think of a particular value that helps solve this problem?

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