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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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3  
To find poles of rational functions, look for zeros of their denominators. To see if a pole is simple, see if it is a simple root of the denominator. –  anon Jul 19 '12 at 2:52
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So your example has simple poles at each of the four 4th roots of -16. –  GEdgar Jul 19 '12 at 2:57
    
I don't understand what a "simple root of the denominator" is. I get that the 4th roots of -16 is a pole. However, I don't understand how it is a simple pole. –  Joebevo Jul 19 '12 at 3:05
    
The denominator of a rational function will be a polynomial. A root $r$ of a polynomial $p(x)$ is a simple root if the linear factor $x-r$ only appears once in its complete factorization. –  anon Jul 19 '12 at 3:13
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Without knowing the complete factorization, you can still check which roots, if any, are simple: a root $r$ of $p(x)$ is simple iff it is not a root of $p'(x)$. So any non-simple roots of $p(x)$ will be roots of $\gcd(p(x),p'(x))$. If the gcd of $p(x)$ and $p'(x)$ is $1$, all roots are simple. –  Robert Israel Jul 19 '12 at 5:20

3 Answers 3

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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Here is how you find the roots of $z^4+16=0$,

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

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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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Thanks, but I don't quite understand how to get from step 2 to step 3, which is important if I want the exact expression for the pole. How exactly did you factor those terms? Is there a general method? –  Joebevo Jul 19 '12 at 3:15
    
it's just the difference of two squares –  Kris Jul 19 '12 at 3:21

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