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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:

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

  • 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.2012-07-19
  • 1
    So your example has simple poles at each of the four 4th roots of -16.2012-07-19
  • 0
    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.2012-07-19
  • 0
    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.2012-07-19
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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.2012-07-19
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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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