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In general, how do I recognize that a simple pole exists and find it, given some $\Large f(z)$? I want to do this without finding the Laurent series first.

And specifically, for the following function:

$\Large f(z) = \frac{z^2}{z^4+16}$

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    In your example, the only root of $(z^4 + 16)' = 4 z^3$ is $0$, which is not a root of $z^4+16$, so all roots of $z^4+16$ are simple.2012-07-19

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Addressing Joebevo's first comment to BruinJ's answer: $z^2-4i=0\Longrightarrow z=\pm \sqrt{4i}=\pm 2\sqrt i=\pm 2\left(\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}i\right)=\pm\sqrt 2(1+i)$ and now you factor easily: $z^2-4i=\left(z-\sqrt 2(1+i)\right)\left(z+\sqrt 2(1+i)\right)$ You can do simmilarly for the other quadratic factor.

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$f(z) = \frac{z^2}{z^4 + 16} = \frac{z^2}{(z^2-4i)(z^2+4i)} = \frac{z^2}{(z\pm (1+i)\sqrt{2})(iz+ (1+i)\sqrt{2})((1+i)\sqrt{2}-iz)} $

Even if you couldn't have factored $(z^2-4i)$ or $(z^2+4i)$, you should be able to see it could factor into unique terms of order 1. The rest is pretty straightforward from your book.

Edit: sorry I totally factored wrong haha. But you get the point heh.

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    it's just the difference of two squares2012-07-19
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Here is how you find the roots of $z^4+16=0$,

$ z^4 = -16 \Rightarrow z^4 = 16\, e^{i\pi} \Rightarrow z^4 = 16\, e^{i\pi+i2k\pi} \Rightarrow z = 2\, e^{\frac{i\pi + i2k\pi}{4} }\, k =0\,,1\,,2\,,3\,.$