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I'm trying to write the function $f(z)= (e^z-1)/z$ when $z \neq 0$ and 1 when $z=0$ in the form of $u(x,y)+iv(x,y)$ where $z=x+iy$. How should I proceed?

I have already shown that it is differentiable (continuously) for all $z \in \mathbb C$ in an earlier problem, but is this needed to solve this?

Also, another problem asks to show that $u(x,y)$ is $C^\infty$; this is simply a result of the fact that $(e^z)'=e^z$ correct? Thank you.

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Pity. $$\frac z {e^z-1}$$ is far mor interesting. –  Pedro Tamaroff May 31 '12 at 23:33
    
Welcome to m.se, Mr President. –  Gerry Myerson Jun 1 '12 at 0:27
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1 Answer

$$\frac{e^z-1}{z}=\frac{e^xe^{iy}-1}{x+iy}=\frac{e^x\cos y-1+ie^x\sin y}{x+iy}$$ and now do the usual to divide two complex numbers.

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Now, the real test... Can our helpful members refrain from doing more until James returns and gives this a try? –  GEdgar Jun 1 '12 at 0:23
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