Residues of the function $\exp\big(\frac{2z}{z^2-1}\big)$ I am trying to calculate the residues of the following analytic function:
$$
f(z)=\exp\Big(\frac{2z}{z^2-1}\Big).
$$
Given that $z=\pm1$ are simple poles for the exponent, and given that the related function $\exp(1/z)$ has an essential singularity in the simple pole of the exponent, I think that I can deduce that the given function has two essential singularities in $z=\pm1$.
Is this reasoning correct?
Anyway, the following limit is indeterminate for any positive $n$
$$
\lim_{z\to1}(z-1)^nf(z),
$$
and this should be enough to say that $z=1$ is an essential singularity, and the same holds for $z=-1$.
Moreover, it is not difficult to find the residue at infinite, where the function is analytic, and I find
$$
R[\infty]=-\lim_{z\to\infty} z\big(f(z)-f(\infty)\big)=-2,
$$
where
$$
f(\infty)\equiv \lim_{z\to\infty} f(z)=1.
$$
Anyway I cannot use the relation
$$
R[+1]+R[-1]+R[\infty]=0,\tag{1}
$$
because there are two residues that I don't know yet.
I am not able to use known series expansions to determine the missing residues, nor I am able to use the definition of the residue as an integral.
This definition by the integral can be used to perform a numerical calculation, for which I obtain
$$
R[+1] = 1.479679824344824\\
R[-1] = 0.520320175655173
$$
and these values confirm the validity of $(1)$.
My question is: there is a way to determine the residues in the two essential singularities in a closed form by exact calculations?
 A: Taking into account the simple fractions decomposition
$$
\frac{2z}{z^2-1}=\frac{1}{z-1}+\frac{1}{z+1}
$$
we have
$$
f(z)=\exp\Big(\frac{1}{z-1}\Big)\exp\Big(\frac{1}{z+1}\Big).
$$
Setting $w=z-1,$ we can study the following function around $w=0$
\begin{align}
g(w)&=\exp\Big(\frac{1}{w}\Big)\exp\Big(\frac{1}{w+2}\Big)=\\
  &=\Big(1+\frac{1}{w}+\frac{1}{2w^2}+\frac{1}{3!w^3}+\ldots\Big)\Big(a_0+a_1w+\frac{a_2}{2}w^2+\frac{a_3}{3!}w^3+\ldots\Big)
\end{align}
where the $a_n$ are the derivatives of $\exp[1/(w+2)]$ in $0$.
The residue is the coefficient of the term $1/w$, and this is
$$
R[1,f]=R[0,g]=a_0+\frac{a_1}{2}+\frac{a_2}{2\cdot3!}+\frac{a_3}{3!\cdot4!}+\ldots=\sum_{n=0}^\infty\frac{a_n}{n!(n+1)!}\tag1
$$
The coefficients $a_n$ are of the form
$$
a_n=b_n\cdot\frac{\sqrt{e}}{(-4)^n}
$$
with the first few $b_n$ given by
$$
1, 1, 5, 37, 361, 4361, 62701,\ldots
$$
that substituted into $(1)$ provide the approximation $1.4796803093686572,$ that could be improved taking more terms.
In the same manner can be evaluate the residue in $z=-1.$
