Use Cauchy's Integral Formula to evaluate the following integral. $$\int \frac{e^z\sin z}{(2z+5i)^2} \, dz,$$ where $$\gamma$$ is a circumference of radius $5$ centered at $-4$ and traversed once in the negative (with respect to its interior) direction.

I've been trying to rework the fraction in the integral to make it suitable for the Cauchy integral formula.  I tried expanding the denominator into $(2z+5i)(2z+5i)$ and then I pulled out some factors from the denominator to get: 
$$ \frac{\frac{e^z\sin z}{2(2z+5i)}}{z-(-\frac{5}{2i})},$$
I suspect that this is not valid because the function in the numerator fails to be analytic at the point $-5/2$ which IS in the domain.  Am I approaching this in the right way? Can anyone help me complete this problem? 
 A: The integrand $f$ has a singularity of order 2 at $z=-\frac52 i$.  To find the residue, use the following for $n=2$ and $z_0=-\frac52$
$$\begin{align}
\text{Res}_{z_0}(f) &= \frac{1}{(n-1)!} \lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}}  \left((z-z_0)^n f(z)\right) \\
&=\frac{1}{(2-1)!} \lim_{z \to -\frac52 i} \frac{d^{2-1}}{dz^{2-1}}  \left((z+\frac52 i)^2 \frac{e^z \sin z}{(2z+5i)^2}\right) \\
&= \lim_{z \to -\frac52 i} \frac{d}{dz}  \left(\frac{e^z \sin z}{4}\right) \\
&=\frac14 \lim_{z \to -\frac52 i} e^z(\sin z +\cos z)\\
&=\frac14 e^{-i\frac52} (\sin (-i\frac52) +\cos(-i\frac52))\\
&=\frac14 \left(\cos(5/2)-i\sin(5/2)\right)\left(\cosh(5/2)-i\sinh(5/2)\right)\\
&=\frac14 \left(\cos(5/2)\cosh(5/2)-\sin(5/2)\sinh(5/2)\right)-\frac{i}{4}\left(\cos(5/2)\sinh(5/2)+\sin(5/2)\cosh(5/2)\right)
\end{align}$$

A second way to find the residue of $f$ is to expand the numerator in a power series at $z=-i\frac52$.  To that end, let $w=z+i\frac52$.  Then, we have
$$\begin{align}
\frac{e^z\sin z}{(2z+i5)^2}&=\frac14\; \frac{e^{-5i/2}e^w \sin(w-i5/2)}{w^2}\\
&=\frac14 \; e^{-5i/2} \frac{(1+w+O(w^2))(\sin(-i5/2)+\cos(i5/2)w+O(w^2))}{w^2}\\
&=\frac14 \; e^{-5i/2} \left(\frac{\sin(-i5/2)}{w^2}+\frac{\sin(-i5/2)+\cos(i5/2)}{w}+O(w)\right)
\end{align}$$
from which the residue is the coefficient on the $w^{-1}$ term, namely
$$\text{Res}_{-i5/2}(f)=\frac14 e^{-5i/2} (\sin(-i5/2)+\cos(i5/2))$$
which is identical to result using the previous approach!
