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Let $f$ be the function defined by $$f(z)= \exp \left( \csc \left( \frac{1}{z} \right) \right).$$ I want to compute the residues of $f$ at all isolated singularities of $f$.

There is an isolated singularity at $z = \frac{1}{k\pi}$, for all $k \in \mathbb{Z}$, but $z=0$ is not an isolated singularity, since $\frac{1}{k\pi} \to 0$ as $k \to \infty$. We can also see that $f$ has an isolated singularity at $z=\infty$ by consider $f(1/z)$.

My problem is computing the residues at these points however, since the Laurent series seems computationally unavailable.

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    $\begingroup$ Please do not just delete your questions once they got an answer. $\endgroup$
    – quid
    Nov 4 '17 at 15:31
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$$e^{\csc(1/z)}=\sum_{n=0}^{\infty}\dfrac{(\csc(1/z))^n}{n!}$$ Therefore, if we can fond the residue of $\csc(1/z)$ at each of its simple pole, then we can use above formula to find what we want. It is not difficult to prove that $$\csc(1/z)=\dfrac{(-1)^{k+1}}{k^2\pi^2(z-1/k\pi)}+\dfrac{(-1)^{k+1}}{k\pi}+(-1)^{k+1}a_1(k)\left(z-\dfrac1{k\pi}\right)+\cdots$$ and therefore $$\text{Res}(\csc(1/z),1/k\pi)=\dfrac{(-1)^{k+1}}{k^2\pi^2}.$$ Then $e^{\csc(1/z)}=$ $$1+\dfrac{1}{1!}\left(\dfrac{(-1)^{k+1}}{k^2\pi^2(z-1/k\pi)}+\dfrac{(-1)^{k+1}}{k\pi}+\cdots\right)+\dfrac{1}{2!}\left(\dfrac{(-1)^{k+1}}{k^2\pi^2(z-1/k\pi)}+\dfrac{(-1)^{k+1}}{k\pi}+\cdots\right)^2+\dfrac{1}{3!}\left(\dfrac{(-1)^{k+1}}{k^2\pi^2(z-1/k\pi)}+\dfrac{(-1)^{k+1}}{k\pi}+\cdots\right)^3+\cdots.$$

From here we can write the residue at each isolated singularity as an infinite series. $$\text{Res}(e^{\csc(1/z)},1/k\pi)=a+ab+\dfrac{ab^2}{2!}+\dfrac{ab^3}{3!}+\cdots=ae^b$$ where $a=\dfrac{(-1)^{k+1}}{k^2\pi^2}$ and $b=\dfrac{(-1)^{k+1}}{k\pi}.$

However I still believe that there is a simple solution than this.

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