Evaluate $\int_{C[0,3]} \frac{\exp(z)}{(z+2)^2\sin(z)} dz$ Using Residue Theorem $\displaystyle \int_{C[0,3]} \frac{\exp(z)}{(z+2)^2\sin(z)} \, dz$ using Residue Theorem.
I have found singularities within $C[0,3]$, which are $-2$ and $0$. For $z=-2$, it is a pole with degree $2$. However, I do not know what kind of singularities $z=0$ is.
 A: To answer the specific issue raised in the OP, the pole at $z=0$ is a simple pole since $\sin(z)=z(1+O(z^2))$ and therefore $\frac{1}{\sin(z)}=\frac{1}{z(1+O(z^2))}=\frac{1}{z}+O(1)$
Now, as stated in the comments, there are a number of ways forward to determining the residue at $z=-2$.  
METHODOLGY $1$:  Apply Standard Limit Formula
First, we can apply the Limit Formula for evaluating the residue at at a pole of order $n$ and write for $n=2$
$$\begin{align}
\text{Res}\left(\frac{e^z}{(z+2)^2\sin(z)}, z=-2\right)&=\lim_{z\to -2}\frac{d}{dz}\left(\frac{e^z}{\sin(z)}\right)\\\\
&\bbox[5px,border:2px solid #C0A000]{=-e^{-2}\frac{1+\cot(2)}{\sin(2)}} \tag 1
\end{align}$$

METHODOLGY $2$:  Find the Coefficient on $\frac{1}{z+2}$ of the Laurent Series
Second, we can determine the coefficient of the $(z+2)^{-1}$ term in the Laurent series for the integrand.  We proceed by expanding the numerator and denominator in their Taylor series to reveal that 
$$\begin{align}
\frac{e^z}{(z+2)^2\sin(2)}&=\frac{e^{-2}\left(1+(z+2)+O(z+2)^2\right)}{(z+2)^2\left(\sin(-2)+\cos(-2)(z+2)+O(z+2)^2\right)}\\\\
&=-\frac{e^{-2}}{\sin(2)}\frac{\left(1+(z+2)+O(z+2)^2\right)\left(1+\cot(2)(z+2)+O(z+2)^2\right)}{(z+2)^2}\\\\
&=-\frac{e^{-2}}{\sin(2)}\left(\frac{1}{(z+2)^2}+\frac{1+\cot(2)}{z+2}+O(1)\right)
\end{align}$$
Inasmuch as the residue is the coefficient on the $\frac{1}{z+2}$ term, we find that 
$$\bbox[5px,border:2px solid #C0A000]{\text{Res}\left(\frac{e^z}{(z+2)^2\sin(z)}, z=-2\right)=-e^{-2}\frac{1+\cot(2)}{\sin(2)}}$$
recovering the result reported in $(1)$!
