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Zeros and poles of rational function.

Mathematics Asked on December 20, 2021

Suppose we are given a rational function
$$f(s) = frac{25s}{s^4+18s^3 +134s^2 +472s+680} $$ and we need to find the zeros and poles of the function.

Suppose $f(s) = frac{a(s)}{b(s)}$, then $a(s) = 25s$ and $s = 0$ is a root of $a(s) =0$ and hence $0$ is a zero of $f(s)$. Again $b(s) = 0$ has roots at $s = -5 pm3i$ and $s = -4pm2i$, hence $-5 pm3i$ and $-4pm2i$ are poles of $f(s)$.

Is the solution correct?

One Answer

Just because $s = 0$ is a zero of $a(s)$ does not mean $s = 0$ is a zero of $f(s)$. To be a zero, the numerator must be zero and the denominator not (otherwise, apply l'Hopital's rule to agonizingly cancel the common factor in $a$ and $b$). Happily, $b(0) neq 0$ and once you observe this fact, you have found all the zeroes of $f$ (and shown that it is a genuine zero of $f$). (Example: $z/z$ has a removable singularity at $z = 0$ and upon removal takes the value $1$.)

$f$ has a pole at $s$ if $b(s) = 0$ and $a(s) neq 0$. With the above modification, you would have already disposed of the possibility of common zeroes in $a$ and $b$, so every zero of $b$ is a pole, as you write.

Answered by Eric Towers on December 20, 2021

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