Contour integral of $f(z) = \frac{1}{z^2+iz+6} $ Need help evaluating a certain contour integral.
$f(z) = \frac{1}{z^2+iz+6}  $
Steps so far:
Poles:  $ z^2+iz+6 \rightarrow \frac{-i \pm \sqrt{-1-24}}{2}=0 \rightarrow z_0 = +2i, -3i $ 
Residues:  $ a_{-1} = \lim{z \to z_{i}} [ \frac{d^{n-1}}{dz^{n-1}} (z-z_{0})^n f(z)]   $
Plug all this in I get:
$\frac{1}{z+3i} +\frac{1}{z-2i} = a_{-1}    $
$ \therefore   \oint \frac{1}{z^2+iz+6}dz  = -2\pi i[\frac{1}{z+3i}+\frac{1}{z-2i} ]$
Have I done something wrong?  
 A: Let $f(z)$ be defined as 
$$\begin{align}
f(z)&=\frac{1}{z^2+iz+6}\\\\
&=\frac{1}{(z+i3)(z-i2)}\\\\
&=\frac{1}{i5}\left(\frac{1}{z-i2}-\frac{1}{z+i3}\right)
\end{align}$$
Therefore, $f(z)$ has poles at $z=-i3$ and $z=i2$.  Integration of $f(z)$ around a closed rectifiable contour $C$ yields
$$\oint_C f(z)\,dz=\begin{cases}
0&,\text{if}\,\,C\,\,\text{does not encircle either pole}\\\\
\frac{2\pi}{5} &,\text{if}\,\,C\,\,\text{encircles only the pole at}\,\,z=i2\\\\
-\frac{2\pi}{5}&,\text{if}\,\,C\,\,\text{encircles only the pole at}\,\,z=-i3\\\\
0&,\text{if}\,\,C\,\,\text{encircles both poles}\\\\
\end{cases}$$

NOTE:
To use the limit definition for evaluating residues, we have for the residue at $z=i2$
$$\begin{align}\text{Res}\left(\frac{1}{z^2+iz+6},z=i2\right)&=\lim_{z\to i2}\frac{z-i2}{z^2+iz+6}\\\\
&=\lim_{z\to i2}\frac{1}{z+i3}\\\\
&=\frac{1}{i5}
\end{align}$$
and for the residue at $z=-i3$
$$\begin{align}\text{Res}\left(\frac{1}{z^2+iz+6},z=-i3\right)&=\lim_{z\to -i3}\frac{z+i3}{z^2+iz+6}\\\\
&=\lim_{z\to -i3}\frac{1}{z-i2}\\\\
&=-\frac{1}{i5}
\end{align}$$
