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If we let $f(z)=(1-e^{1/z})^{-1} ,$ where $z$ is complex, I'm trying to use the Cauchy-Riemann equations to determine if $f$ is holomorphic. So I need to separate it into real and imaginary functions of $x$ and $y$ ($u(x,y)$ , $v(x,y)$ respectively), then partially differentiate.

So far I have

$u(x,y)=\frac {1-e^{\frac {x}{x^{2}+y^{2}}} \cos(\frac{y}{x^{2}+y^{2}})}{1-2e^{x} \cos(\frac{y}{x^{2}+y^{2}})+e^{\frac {1}{x^{2}+y^{2}}}} $

$v(x,y)=\frac {-e^{\frac {x}{x^{2}+y^{2}}} \sin(\frac{y}{x^{2}+y^{2}})}{1-2e^{x} \cos(\frac{y}{x^{2}+y^{2}})+e^{\frac {1}{x^{2}+y^{2}}}} $

This looks a bit of a nightmare to differentiate 4 times for a small part of a question, can anyone see how a quicker way to use the C-R equations to determine if $f$ is holomorphic?

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    @ThomasRot: Yes indeed, so it is certainly not holomorphic on the whole complex plane2011-09-20

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The OP got the answer from the hints. I'll just post it here so this can have an accepted answer.

Compositions of holomorphic functions are holomorphic. Because the functions $z\mapsto 1-z$, $z\mapsto e^z$, and $z\mapsto 1/z$ are holomorphic (outside z=0 for the latter case), we find that the function $f$ is holomorphic, whenever it is defined. This is for $z\not =0$ and $z\not= \frac{i}{2\pi n}$ with $n\not =0$ an integer.