Vrftsjtryk

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}{2pi i}oint_C frac{f(z)dz}{(z-z_0)^{n+1}}\
% && text{Regular Part}&\
% b_n &= frac{1}{2pi 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}{2pi i}oint_C frac{frac{1}{(z+2i)^3(z-2i)^3}}{(z-2i)^{n+1}}\
a_n &= frac{1}{2pi 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}{2pi i}oint_C frac{1}{(z+2i)^{3}(z-2i)^{-1+4}}\
&=frac{1}{2pi i}oint_C frac{1}{(z+2i)^{3}(z-2i)^{3}}\
&=frac{1}{2pi 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 ;)

Integration is completely unnecessary here. More generally, never integrate if you can differentiate. And most of the times, geometric series is more than enough.

$$
begin{aligned}
frac{1}{(z+2i)^3}&=frac{1}{2}frac{mathrm d^2}{mathrm dz^2}frac{1}{(z-2i)+4i}\
&=frac12frac{mathrm d^2}{mathrm dz^2}left[-frac i4+frac{1}{16}(z-2i)+frac{i}{64}(z-2i)^2+cdotsright]\
&=frac{i}{64}-frac{3}{256}(z-2i)-frac{3i}{512}(z-2i)^2+cdots
end{aligned}
$$

Therefore,
$$
begin{aligned}
f(x)&=frac{1}{(z-2i)^3}left[frac{i}{64}-frac{3}{256}(z-2i)-frac{3i}{512}(z-2i)^2+cdotsright]\
&=frac{i}{64}(z-2i)^{-3}-frac{3}{256}(z-2i)^{-2}color{red}{-frac{3i}{512}(z-2i)^{-1}}+cdots
end{aligned}
$$

Using the formula
$$
frac{1}{1-z}=sum_{nge0}z^n
$$
you should be able to write down a general formula for $a_n$, using the steps described above. I leave this to you.

Here's my final answer (I use the physics package in Latex, sorry for the weird formatting in some areas):
begin{align}
f(z) &= frac{1}{(z^2+4)^3}\
&= frac{1}{(z-2i)^3}frac{1}{(z+2i)^3}
end{align}
Here, we see third order poles at $z=pm2i$. The Laurent series expansion of $f(z)$ about $z=2i$ should have the form,
begin{align}
f(z) &= frac{a_{-3}}{(z-z_0)^{-3}}+ frac{a_{-2}}{(z-z_0)^{-2}} + frac{a_{-1}}{(z-z_0)} + a_0 + a_1(z-z_0) + a_2(z-z_0)^2+cdots\
end{align}
To create an expandable geometric series for the $a_{-3}$ term, the following procedure can be used,
begin{align}
(z-z_0)^{3}f(z) &= a_{-3}+{a_{-2}}{(z-z_0)}+a_{-1}(z-z_0)^2 + a_0(z-z_0)^3 + a_1(z-z_0)^4 + a_2(z-z_0)^5+cdots
end{align}
Differentiating both sides, we find
begin{align}
frac{d^2}{dz^2}[(z-z_0)^{3}f(z)] &= (2cdot1)a_{-1} + sum_n^infty b_n(z-z_0)^n\
end{align}
and in the limit where $zto z_0$,
begin{align}
a_{-1}& = frac{1}{2}frac{d^2}{dz^2}[(z-z_0)^{3}f(z)]
end{align}
which, of course, is the residue of the function at $z=2i$. Therefore,
begin{align}
f(z) &= frac{1}{(z-2i)^3}frac{1}{2}frac{d^2}{dz^2}(frac{1}{(z+2i)})\
&= frac{1}{(z-2i)^3}frac{1}{2}frac{d^2}{dz^2}(
frac{1/4i}{(1+frac{z-2i}{4i})})\
%
&= frac{1}{(z-2i)^3}frac{1}{8i}frac{d^2}{dz^2}[
1-frac{z-2i}{4i}
+(frac{z-2i}{4i})^2-(frac{z-2i}{4i})^3
+(frac{z-2i}{4i})^4-(frac{z-2i}{4i})^5
+cdots]\
%
&= frac{1}{(z-2i)^3}frac{1}{8i}frac{d}{dz}[
-frac{1}{4i}
+frac{2(z-2i)}{(4i)^2}-frac{3(z-2i)^2}{(4i)^3}
+frac{4(z-2i)^3}{(4i)^4}-frac{5(z-2i)^4}{(4i)^5}
+cdots]\
%
&= frac{1}{(z-2i)^3}frac{1}{8i}[
frac{2cdot1}{(4i)^2}-frac{3cdot2(z-2i)}{(4i)^3}
+frac{4cdot3(z-2i)^2}{(4i)^4}-frac{5cdot4(z-2i)^3}{(4i)^5}
+cdots]\
%
&= frac{1}{(z-2i)^3}frac{1}{8i}[
frac{2cdot1}{(4i)^2}-frac{3cdot2(z-2i)}{(4i)^3}
+frac{4cdot3(z-2i)^2}{(4i)^4}-frac{5cdot4(z-2i)^3}{(4i)^5}
+cdots]
end{align}
After simplifying, we find the residue is $-3i/512$ and,
begin{equation}
boxed{
f(z) = frac{i}{64}(z-2i)^{-3}-frac{3}{256}(z-2i)^{-2}
-frac{3i}{512}(z-2i)^{-1}+frac{5}{2048}+cdots
}
end{equation}