Contour Integration cauchy integration formula If I have the integrand 

$$\frac{e{^z}^2}{(2z+i)(z+3i)^2}$$ 

for the circles with centre $a$ radius $2$.
I know the integrand has a single pole at  $z=-i/2$ and a double pole at $z=-3i$
My question is for centres $a=3,5,7i$ how do I know if the poles lie inside the circle to enable use of the Cauchy Integral formula, I'm getting confused as there are $i$ terms. Thank you.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{a \in \braces{3,5,7\ic}}$.

\begin{align}
&\bbox[10px,#ffd]{\ds{\oint_{\verts{z - a}\ =\ 2}{\expo{z^{2}} \over
\pars{2z + \ic}\pars{z + 3\ic}^{2}}\,\dd z}} =
{1 \over 2}\expo{a^{2}}\oint_{\verts{z}\ =\ 2}{\expo{z^{2} + 2az} \over
\bracks{z - \pars{-a - \ic/2}}\bracks{z -\pars{-a - 3\ic}}^{\, 2}}\,\dd z
\end{align}

$\ds{a \in \braces{3,5,7\ic} \implies \verts{-a - {\ic \over 2}} > 2}$ and
  $\ds{\verts{-a - 3\ic} > 2}$. So, the above integral
  $\ds{\color{red}{\texttt{vanishes out}}}$.

