Finding Residue I am having difficulty with with calculating the residue for $$\text{res}[\frac{\exp(\frac{1}{z})}{z^{2}-16,},z=0]$$ 
 I was able to calculate the residues when $z=4$ and $z=-4$. However Im not sure how to approach this part of the question.
 A: Let $f(z) = \frac{e^{1/z}}{z^2 - 16}$. Let $C_R$ be the circle of radius R centered at the origin.  For $R > 4$, 
$$
Res_{z=0}f(z) + Res_{z=4}f(z) + Res_{z = -4}f(z) = \frac{1}{2\pi i}\int_{C_R}f(z)dz
$$
Note that 
$$
|\frac{1}{2\pi i}\int_{C_R} \frac{e^{1/z}}{z^2-16} dz| = |\frac{1}{2\pi i} \int_{0}^{2\pi} \frac{e^{(1/R) e^{-i\theta}}}{R^2e^{2i\theta} - 16}iRe^{i\theta} d\theta| \leq \frac{1}{2\pi}\int_0^{2\pi}\frac{R }{R^2 - 16}e^{(1/R)cos\theta} d\theta 
$$
$$
\leq \frac{1}{2\pi}\frac{R}{R^2 - 16} \int_{0}^{2\pi} e^{(1/R)} d\theta.
$$
Letting $R \to \infty$, this integral goes to 0. Thus,
$$
Res_{z=0}f(z) = - Res_{z=4}f(z) - Res_{z = -4}f(z).
$$
Since you said you computed the right hand side, you now have the answer.
A: $e^z$ is an entire function, so $z = 0$ is an essential singularity for $e^{1/z}$.
$$e^{1/z} = \sum_{k = 0}^{+\infty} \frac{1}{k!}(z^{-k})$$
$$\frac{1}{z^2-16} = \frac{1}{16\left(\frac{z^2}{16} - 1\right)} = \frac{-1}{16} \sum_{n = 0}^{+\infty} \frac{z^{2n}}{16^n}$$
You can see the first terms of the expansion ad $z = 0$:
$$-\frac{1}{16}\left(1 + \frac{1}{z} + \frac{1}{2z^2} + \frac{1}{3!z^3} + \frac{1}{4!z^4} + \frac{1}{5!z^5} + \frac{1}{6!z^6} + \frac{1}{7!z^7} + \frac{1}{8!z^8} + \frac{1}{9!z^9} + \ldots   \right) \times \\\\ \times \left(1 + \frac{z^2}{16} + \frac{z^4}{16^2} + \frac{z^6}{16^3} + \frac{z^8}{16^4} + \ldots\right)$$
Now you know that the residue is the coefficient of the term $z^{-1}$ and you can easily obtain it by finding all the term depending only on $z^{-1}$: for example the second terms on the left times the first term on the right bracket, plus the the fourth term on the left bracket times the second on the right bracket and so on, and you will get
$$-\frac{1}{16}\left(\frac{1}{z} + \frac{1}{16\times 3! z} + \frac{1}{16^2\times 5! z} + \frac{1}{16^3\times 7! z} + \ldots\right)$$
It's easy to guess the series for that expansion:
$$\frac{-1}{16z} \sum_{k = 0}^{+\infty} \frac{1}{16^k\cdot (2k + 1)!}$$
The series converges to 
$$4\sinh \left(\frac{1}{4}\right)$$
Hence the residue is
$$\frac{-1}{4} \sinh \left(\frac{1}{4}\right)$$
That is indeed the value of your residue
$$-0.06315307920204207...$$
