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?
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 $z=-i3$
$$\begin{align}\text{Res}\left(\frac{1}{z^2+iz+6},z=-i3\right)&=\lim_{z\to -i3}\frac{z+i3}{z^2+iz+6}\\\\ &=\lim_{z\to -i3}\frac{1}{z-i2}\\\\ &=-\frac{1}{i5} \end{align}$$