<s>[INST] Residue of order 3 - Find the Laurent Series for the function
  \begin{align}
    f(z) = \frac{1}{(z^2+4)^3}
  \end{align}
  about the isolated singular pole $z = 2i$. What is the pole order? What is the residue at the pole?
My attempt:
  \begin{align}
    f(z) &= \frac{1}{(z^2+4)^3}\\
      &= \frac{1}{(z+2i)^3(z-2i)^3}\\
  \end{align}
  Here we see $z=2i$ is a 3rd order pole.
A Laurent series is defined with respect to a particular non-analytic point $z_0$ and a path of integration C. The path of integration must lie in an annulus surrounding $z_0$ and so
   \begin{align}
     f(z) &= \sum_{n=-\infty}^\infty a_n(z-z_0)^n
     % = \sum_{n=0}^\infty a_n(z-z_0)^{n}
     % + \sum_{n=1}^\infty b_n(z-z_0)^{n}
   \end{align}
   where
   \begin{align}
     a_n &= \frac{1}{2\pi i}\oint_C \frac{f(z)dz}{(z-z_0)^{n+1}}\\
     % &&  \text{Regular Part}&\\
     % b_n &= \frac{1}{2\pi i}\oint_C \frac{f(z)dz}{(z-z_0)^{-n-1}} \label{eq:laurentb}
     % && \text{Principle Part} &
   \end{align}
   We find the $a_n$ term using $z_0=2i$,
   \begin{align}
     a_n &= \frac{1}{2\pi i}\oint_C \frac{\frac{1}{(z+2i)^3(z-2i)^3}}{(z-2i)^{n+1}}\\
     a_n &= \frac{1}{2\pi i}\oint_C \frac{1}{(z+2i)^3(z-2i)^{n+4}}
   \end{align}
   and from Cauchy's Integral Formula we can find the residue $(n=-1)$ term,
   \begin{align}
     a_{(-1)}&=\frac{1}{2\pi i}\oint_C \frac{1}{(z+2i)^{3}(z-2i)^{-1+4}}\\
     &=\frac{1}{2\pi i}\oint_C \frac{1}{(z+2i)^{3}(z-2i)^{3}}\\
     &=\frac{1}{2\pi i}\oint_C \frac{1}{(z+2i)^{3}(\sqrt{z}+\sqrt{2i})}
      \frac{1}{(z-2i)(\sqrt{z}-\sqrt{2i})}????
   \end{align}
I can't seem to get anywhere near a correct answer ($Res(f;2i) = -3i/512$).  I'm supposed to use the Laurent expansion at $z=2i$ and ultimately find the $z^-1$ coefficient, but I'm so lost.... I've also tried partial fraction expansion, but am running in circles.  Engineering student here, so be nice ;)
 [/INST] Integration is completely unnecessary here. More generally, never integrate if you can differentiate. And most of the times, geometric series is more than enough.
$$
\begin{aligned}
\frac{1}{(z+2i)^3}&=\frac{1}{2}\frac{\mathrm d^2}{\mathrm dz^2}\frac{1}{(z-2i)+4i}\\
&=\frac12\frac{\mathrm d^2}{\mathrm dz^2}\left[-\frac i4+\frac{1}{16}(z-2i)+\frac{i}{64}(z-2i)^2+\cdots\right]\\
&=\frac{i}{64}-\frac{3}{256}(z-2i)-\frac{3i}{512}(z-2i)^2+\cdots
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
f(x)&=\frac{1}{(z-2i)^3}\left[\frac{i}{64}-\frac{3}{256}(z-2i)-\frac{3i}{512}(z-2i)^2+\cdots\right]\\
&=\frac{i}{64}(z-2i)^{-3}-\frac{3}{256}(z-2i)^{-2}\color{red}{-\frac{3i}{512}(z-2i)^{-1}}+\cdots
\end{aligned}
$$
Using the formula
$$
\frac{1}{1-z}=\sum_{n\ge0}z^n
$$
you should be able to write down a general formula for $a_n$, using the steps described above. I leave this to you.
</s>