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I've literally tried every technique I know of and they all lead to explosions of the kind $1/0$. Generally speaking the residue at $c$ for a function can be calculated as:

$\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\Big((z-c)^n f(z) \Big)$ where $n$ is the order of the pole or higher.

Naturally I did the case $n=2$ as this seems like a 2nd order pole, but it still diverges. I tried higher numbers, I even entered $n=20$ in Mathematica to no avail.

What am I doing wrong? Is there any trick I should see here? I don't think so since the formula is quite general.

Also, note that there does exist a residue and it is $\frac{3i}{32}$

DLV
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    Have you tried partial fractions? – Thomas Andrews Sep 05 '15 at 01:09
  • BEHOLD: http://www.wolframalpha.com/input/?i=residue+of+\frac{1}{z%28z^2%2B4%29%28z%2B2i%29}+at+z%3D-2i which is computable from http://www.wolframalpha.com/input/?i=partial+fractions+of+\frac{1}{z%28z^2%2B4%29%28z%2B2i%29} you have to cut and paste the links not just click them for full-effect. – James S. Cook Sep 05 '15 at 01:13
  • Your $f(z)$ should have been $1/z(z-2i)$ and $n=2$. Maybe you mistook the $z-2i$ for $z+2i$. – James S. Cook Sep 05 '15 at 01:18

2 Answers2

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Partial fractions: $$\frac{1}{z(z^2+4)(z+2i)}=\frac{a}{z}+\frac{b}{z-2i}+\frac{c}{z+2i}+\frac{d}{(z+2i)^2}$$

You need to find $c$.

Thomas Andrews
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$$(z+2i)^2f(z)=\frac{1}{z(z-2i)} $$

$$\frac{d}{dz} \frac{1}{z(z-2i)} = -\frac{2z-2i}{z^2(z-2i)^2}$$

evaluating at $z=-2i$

$$R=-\frac{-4i-2i}{(-2i)^2(-4i)^2}=-\frac{-6i}{ (-4)(-16)}=\frac{3i}{32}$$

WW1
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