Contour Integration cauchy integration formula

Question

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.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

\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 $\ds{\color{red}{\texttt{vanishes out}}}$.