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Locate the singular points and state whether it is a pole, a removable singularity, or an essential singulatity: $$f(z) = \frac{z}{e^z - 1}.$$

I obtained $z=0$. But I don't understand how to check what kind of singularity it is.

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up vote 2 down vote accepted

When $z=0$, BOTH the numerator and denominator are $0$, and when that happens, it's most often a removable singularity. One way to show that that happens in this case is to expand $e^z$ as a power series and then do two obvious cancelations. Another is to apply L'Hopital's rule to $\lim\limits_{z\to0}f(z)$.

At what points is $e^z-1$ equal to $0$ when $z\ne0$? Poles occur at such places. Remember that if $z=x+iy$ then $e^z=e^{x+iy}$ $=e^x e^{iy}$ $= e^x(\cos y + i\sin y)$. If $x$ and $y$ are real then $e^x$ is positive and $|\cos y +i \sin y|=1$. So the problem is: for what values of $x$ and $y$ is $e^x(\cos y + i\sin y)$ equal to $1$?

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Hint: does $f(z)$ have a limit as $z \to 0$?

Further hint: don't forget there are other values of $z$ for which $e^z = 1$.

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