Finding singularities of a complex function and the respective residues Consider the function, 
$$f(z) = \frac{\sin(\frac{1}{z})}{z^2+a^2} \text{ where } a>0$$
I know that this has two simple poles at $z = \pm ia$ and an essential singularity at $z = 0$ because of $\sin(\frac{1}{z})$ in the numerator. For $\operatorname{Res}(f,ia)$ and $\operatorname{Res}(f,-ia)$, it is relatively easy to find the these values using 
$$a_{-1} = \lim_{z\rightarrow z_0} ((z-z_0)f(z))$$
However, I am finding it much more difficult to determine the residue at an essential singularity.  From what I have read, it seems that the easiest way to determine the residue is to explicitly find the Laurent Expansion and then get it from there. Would this be the case for this function as well?
 A: For positive numers $a<r<R$, Cauchy's Integral Theorem guarantees that  
$$\oint_{|z|=r>a}\frac{\sin(1/z)}{z^2+a^2}\,dz=\oint_{|z|=R>r}\frac{\sin(1/z)}{z^2+a^2}\,dz$$
Then, noting that $\left|\sin\left(\frac1{Re^{i\phi}}\right)\right|\le \frac{R}{R-1}$ for $R>\max(1,a)$
$$\begin{align}
\left|\oint_{|z|=R}\frac{\sin(1/z)}{z^2+a^2}\,dz\right|&=\left|\int_0^{2\pi}\frac{\sin\left(\frac1{Re^{i\phi}}\right)}{R^2e^{i2\phi}+a^2}\,iRe^{i\phi}\,d\phi\right|\\\\
&\le  \int_0^{2\pi} \frac{\left|\sin\left(\frac1{Re^{i\phi}}\right)\right|}{\left|R^2e^{i2\phi}+a^2\right|}\,R\,d\phi\\\\
&\le \frac{R^2}{(R^2-a^2)(R-1)}\\\\
&\to 0\,\,\text{as}\,\,R\to \infty
\end{align}$$
Hence, the sum of the residues of $\frac{\sin\left(1/z\right)}{z^2+a^2}$ is $0$ and we find that 
$$\begin{align}
\text{Res}\left(\frac{\sin\left(1/z\right)}{z^2+a^2},z=0\right)&=-\text{Res}\left(\frac{\sin\left(1/z\right)}{z^2+a^2},z=ia\right)-\text{Res}\left(\frac{\sin\left(1/z\right)}{z^2+a^2},z=-ia\right)\\\\
&=\frac{\sinh(1/a)}{a}
\end{align}$$
