Why is $0$ a pole of first order of $\frac{1}{e^z-1}$?

Question

I am working on the following exercise:

Classify the isolated singularity $0$ of $\frac{1}{e^z-1}$.

I know that $0$ is a pole of first order, but I am struggling to prove it. My attempt goes as follows:

Using the power series-definition of $e^x$ it is easy to see that:

$$\frac{1}{e^z-1} = \frac{1}{z} \cdot \biggl(1 + \sum_{n=1}^{\infty} \frac{z^n}{(n+1)!} \biggr)^{-1}$$

According to my definition for $0$ to be a pole of first order it has to be a removable singularity of the function $$z\cdot \frac{1}{z} \biggl(1 + \sum_{n=1}^{\infty} \frac{z^n}{(n+1)!} \biggr)^{-1} = \biggl(1 + \sum_{n=1}^{\infty} \frac{z^n}{(n+1)!} \biggr)^{-1}.$$

But I do not see why $0$ should be removable in $\biggl(1 + \sum_{n=1}^{\infty} \frac{z^n}{(n+1)!} \biggr)^{-1}$. Could you help me?

Plug in $z=0$ and you get $(1)^{-1}=1$ but you have assumed that $z\cdot\frac1z=1$ i.e. the singularity is removable. — Peter Foreman, Mar 05 '20 at 20:30
Plugging in $z=0$ into the last expression results in no problems. That's why it is a removable singularity, that expression is the analytic function you can replace the original with to be continuous at $z=0$ — Ninad Munshi, Mar 05 '20 at 20:30

score 2 · Accepted Answer · edited Mar 05 '20 at 20:34

2

Given:

$$ g(z) = e^z-1 $$

You can simply notice two things:

$g(0)=0$. This is trivial to assert that it is indeed a pole.
The limit: $$ \lim_{z\to0}\frac{g(z)}{z} = 1 $$

Is finite and non-zero, hence 1 is the multiplicity of the root of $g$ (hence the multiplicity of your pole). You can compute the limit with L'Hospital's rule.

edited Mar 05 '20 at 20:34

Bernard

175,478

answered Mar 05 '20 at 20:32

Mefitico

1,845

Why is $0$ a pole of first order of $\frac{1}{e^z-1}$?

1 Answers1