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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 ;)

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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.

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  • $\begingroup$ Thanks for this. I assume the initial step is from the residue formula $$\operatorname{Res}(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}} \left( (z-c)^n f(z) \right)$$ $\endgroup$ – niagarajohn Nov 17 '18 at 2:14
  • $\begingroup$ Why is it that we expand the $\frac{1}{(z+2i)^3}$ term rather than the other? $\endgroup$ – niagarajohn Nov 17 '18 at 2:28
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    $\begingroup$ I see now, using the fact that $$ \frac{\operatorname{d}^k}{\operatorname{d}\!z^k}\left(\frac{1}{1-z}\right) = \frac{k!}{(1-z)^{k+1}}$$. Thanks again @AccidentalFourierTransform $\endgroup$ – niagarajohn Nov 17 '18 at 4:09
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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)] &= (2\cdot1)a_{-1} + \sum_n^\infty b_n(z-z_0)^n\\ \end{align} and in the limit where $z\to z_0$, \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{2\cdot1}{(4i)^2}-\frac{3\cdot2(z-2i)}{(4i)^3} +\frac{4\cdot3(z-2i)^2}{(4i)^4}-\frac{5\cdot4(z-2i)^3}{(4i)^5} +\cdots]\\ % &= \frac{1}{(z-2i)^3}\frac{1}{8i}[ \frac{2\cdot1}{(4i)^2}-\frac{3\cdot2(z-2i)}{(4i)^3} +\frac{4\cdot3(z-2i)^2}{(4i)^4}-\frac{5\cdot4(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}

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