How to deal with the essential singularity when evaluating $\oint_{C} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{z^2 + 9}dz$ I am trying to use the Cauchy integral formula and the residue theorem to show that
$$
\oint_{|z-i| = 3} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{z^2 + 9} dz= 2\pi i\cosh(1) - \pi i \cosh\left( \dfrac{9+6i}{13}\right)
$$
In preparation to use the Cauchy integral formula, my first attempt was to rewrite the integral:
$$  
\oint_{C} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{(z-3i)(z+3i)}dz = \oint_{C} \dfrac{\dfrac{z\cosh{\left( \frac{z}{z-2}\right)}}{z+3i}} {z-3i}dz
$$
However, the function $g(z) = \frac{z\cosh{\left( \frac{z}{z-2}\right)}}{z+3i}$ fails to be analytic at $z = 2$ which is an interior point of $C = \{z \in \mathbb C : |z-i| = 3  \}$. Therefore, Cauchy's integral theorem does not directly apply. Next, I decomposed one of the multiples into partial fractions:
$$
\dfrac{z}{\left(z - 3 i \right) \left(z + 3 i\right)}=\dfrac{1}{6 i + 2 z} - \dfrac{1}{6 i - 2 z} = \dfrac12\left[ \dfrac{1}{z + 3 i } - \dfrac{1}{3 i - z} \right]
$$
Which gives
$$
\oint_{C} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{(z-3i)(z+3i)}dz = \dfrac12\left[ \oint_C \dfrac{\cosh{\left( \frac{z}{z-2}\right)}}{3 i + z} - \oint_C \dfrac{\cosh{\left( \frac{z}{z-2}\right)}}{3 i - z} \right]
$$
This, unfortunately, does not help with the singularity at $z=2$. In addition, the point $z=3i$ is not even in the region of integration. Thus, I decided to try and solve this using the residue theorem. There are three singular points for the initial function: $\{3i, -3i, 2\}$. Since $-3i \notin C$, we only need the residues at the remaining two points.
Residue at $z=3i$:
$$
\begin{align}
\lim_{z \to 3i} (z-3i)f(z) &= \lim_{z \to 3i} \dfrac{z^2 - 3zi}{z^2 + 9} \cosh\left(\frac{z}{z-2}\right)\\ &= \cosh\left(\frac{3i}{3i-2}\right)\lim_{z \to 3i} \dfrac{z^2 - 3zi}{z^2 + 9}\\ &=\cosh\left(\frac{3i}{3i-2}\right) \cdot \dfrac12
\end{align}
$$
I evaluated the last limit using L'Hopital's rule.
Residue at $z=2$:
The problem is that the limit
$$\lim_{z \to 2} (z-2)f(z)$$
does not exist since the left-hand limit is $-\infty$ whereas the right-hand limit is $+\infty$. Therefore, I cannot use the residue theorem. I am unsure what to try next. How do I deal with the singularity at $z=2$ and what other approaches can I try?
 A: Some ideas for you to evaluate the residue at $\;z=2\;$ of $\;\cosh\frac z{z-2}\;$ (Check carefully the calculations, though the algorithm is the important thing here: sometimes we need, and some even would say must, calculate Laurent series) :
Since $\;\cosh z=\cos(iz)\;$, we get that
$$\cosh z=\sum_{n=0}^\infty \frac{ z^{2n}}{(2n)!}\implies \cosh\frac z{z-2}=\sum_{n=0}^\infty \frac{ \left(\frac z{z-2}\right)^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{  z^{2n}}{(2n)!}\cdot\frac1{(z-2)^n}$$
But
$$z=2+(z-2)=2\left(1+\frac{z-2}2\right)\implies\cosh z=\sum_{n=0}^\infty \frac{2^n\left(1+\frac{z-2}2\right)^n}{(2n)!}\cdot\frac1{(z-2)^n}$$
Let us try to calculate the coefficient $\;c_{-1}\;$  of $\;(z-2)^{-1}\;$ , which is the residue of $\;\cosh\frac z{z-2}\;$ at $\;z=2\;$ . First we write
$$\left(1+\frac{z-2}2\right)^n\cdot\frac1{(z-2)^n}=\left(\frac1{z-2}+\frac12\right)^n=\sum_{k=0}^n\binom nk(z-2)^{-k}\cdot 2^{-n+k}$$
Thus, for each $\;n=1,2,....\;$, the coefficient of $\;(z-2)^{-1}\;$ is given when $\;k=1\;$ (not for $\;n=0\;$...!), and thus it is given by
$$\sum_{n=1}^\infty\binom n1\frac1{2^{n-1}}=\sum_{n=1}^\infty\frac{n}{2^{n-1}}=4\;$$
For the last series' value: differentiate $\;\cfrac1{1-z}\;$ and substitute the correct value of $\;z$
