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Determine the character of the singularity at z=0 for each of the following functions:

a) $f(z)=\frac{1/z^{9}}{e^z-1} $

b) $h(z)=\frac{e^{1/z}-1}{z^{3}} $

c) $g(z)=\frac{e^{z^{9}}-1}{z} $

My attempt

a) For f(z), we have singularity at z=0, both for $\frac{1}{z^9}$ and $\frac{1}{e^z-1}.$

Since $\lim\limits_{z\to 0}\frac{1}{z^9}=\infty$, it must be a pole.

The same goes for $\frac{1}{e^z-1}$, which has simple poles at $z=2\pi i n,\quad n\in0,\pm1,\pm2,...$

b) $h(z)=\frac{e^{1/z}-1}{z^{3}} =\frac{\big(1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+...\big)-1}{z^3}=\frac{1}{z^4}+\frac{1}{2z^5}+...$

So, it's also characterized as a pole (?)

c) $g(z)=\frac{e^{z^{9}}-1}{z} =\frac{\big(1+9z^9+\frac{18z^19}{2!}+...\big)-1}{z}=9z^8+9z^18+....$

Thus, it's a removable singularity.

Is my way of thinking right? If not, what am I doing wrong?

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    If you write "a pole", shouldn't you write down its order? – 2017-02-27
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    I'm suppose to fill out blanks with R (removable), P (poles) or E (essential). So its order is irrelevant in this case :) – 2017-02-27

1 Answers 1

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(a) and (c) are correct but (b) is not.

We call something a pole only when the part of the Laurent series with negative powers of $z$ terminates. Said differently, if $f(z)$ has a pole of order $n$ at $z = 0$, then $z^nf(z)$ should be regular (and non-vanishing) at $z = 0$.

But something like $e^{1/z}$ does not satisfy this property. The Laurent series for this example is $$e^{1/z} = 1 + z^{-1} + \frac 1 {2!} z^{-2} + \frac 1 {3!} z^{-3} + ...$$ The series continues for arbitrarily negative powers of $z$.

We call something like this an essential singularity.

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    Thank you so much. Im still trying to get the hang of this :) – 2017-02-27