# What is the residue of this essential singularity?

I have this function $$f(z) = \frac{e^{1/z}}{ 1 - z^2}$$ which has two poles of order $1$ with residues $- e / 2$ and $1 / (2e)$. Now I need to compute the residue at the essential singularity $z = 0$ where the numerator becomes undefined. I expanded this in series: \begin{align*} f(z) = \frac{1}{- (z + 1) (z - 1)} \big( 1 + \frac{1}{z} + \frac{1}{2} \frac{1}{z^2} + \ldots \big). \end{align*} But I'm not sure how to find the residue from this.

You can easily compute the sum of all three residues as

$$\frac{1}{2\pi i} \int_{\lvert z\rvert = R} \frac{e^{1/z}}{1-z^2}\,dz$$

with $R > 1$. Then just subtract the residues at the poles from the sum to find the residue at $0$.

• Just in case: the residue is $\sinh 1$. – Daniel Fischer Jan 3 '16 at 21:17
• Ok, thanks. Did you use the residue theorem in this case? Evaluation of integral is $2 \pi i$ times the sum of all the residues? – Kamil Jan 3 '16 at 21:19
• Yes. And we can evaluate the integral since it doesn't depend on $R$ (as long as $R > 1$), and the ML inequality suffices. – Daniel Fischer Jan 3 '16 at 21:22
• @CWL The standard estimate yields $$\Biggl\lvert \int_{\lvert z\rvert = R} \frac{e^{1/z}}{1-z^2}\,dz\Biggr\rvert \leqslant 2\pi R \cdot \frac{e^{1/R}}{R^2-1}$$ for $R > 1$, and letting $R \to \infty$, we see that the integral is $0$. – Daniel Fischer May 28 '17 at 14:02
• @CWL To compute $$\int_{\lvert z\rvert = R} \frac{dz}{1-e^{1/z}},$$ make the substitution $w = 1/z$. You get something with only finitely many poles in the interior of the contour. – Daniel Fischer May 28 '17 at 14:10

HINT:

We can write for $|z|<1$

$$e^{1/z}=\sum_{n=1}^\infty \frac{z^{-n}}{n!}$$

and

\begin{align} \frac{1}{1-z^2}&=\sum_{n=0}^\infty z^{2n} \end{align}

Therefore, we have

\begin{align} \frac{e^{1/z}}{1-z^2}&=\sum_{n=1}^\infty \sum_{m=0}^\infty \frac{z^{2m-n}}{n!} \end{align}

The coefficient on $z^{-1}$ comes from the terms for which $2m=n-1$ and therefore the only terms implicated are those for which $n$ is odd. The residue is thus $\sum_{n=1}\frac{1}{(2n-1)!}=\sinh(1)$

The approach of Daniel Fischer is clever. However, maybe you want a direct approach. We can use the expansion $$\frac{e^{1/z}}{1-z^2} = \sum_{j=0}^\infty z^{2j} \sum_{k=0}^\infty \frac{z^{-k}}{k!}$$ valid for $|z|<1$. We need to obtain the residue as the coefficient in from of $z^{-1}$. For that we replace $k$ by the new summation variables $l=2j-k$ with the result $$\frac{e^{1/z}}{1-z^2} = \sum_{l\in\mathbb{Z}}\sum_{j=\max(0,l)}^\infty \frac{z^l}{(2j-l)!} .$$

The residue is thus given by $$\operatorname{Res}_{z=0} \frac{e^{1/z}}{1-z^2} = \sum_{j=0}^\infty \frac1{(2j+1)!}= \sinh(1).$$

• That regrouping of terms is ok only, if both series converge absolutely. They do so in any disk $\delta<|z|<1$ with $\delta>0$. Is that enough? – Jyrki Lahtonen Jan 3 '16 at 21:25
• @JyrkiLahtonen: yes. The series is a Laurent series defined on the ring $0<|z|<1$. The residue is defined as the coefficient in front of $z^{-1}$ of this series. – Fabian Jan 3 '16 at 21:26
• Raking my brain for the correct argument :-). Yeah, I only needed that $\delta$ for uniform convergence. Absolute convergence is fine with $0<|z|<1$. +1 – Jyrki Lahtonen Jan 3 '16 at 21:32