Consider the function $$f(z) := \frac{1}{z^{2n}+1} \quad .$$ The denominator is equal to zero when $z^{2n} = -1$, so its zeros $z_{k}$ are located at $z_{k} = e^{\frac{\pi i}{2n}+\frac{k \pi i}{n}} $. These are the poles of $f$.

I would like to know what the order of these poles is, and how to calculate them. I can't figure it out by means of the definition of a pole of order $m$.

I suppose the poles are of order $1$, since they're all located on a different place on the unit circle.

  • $\begingroup$ All the zeroes of $g(z)=z^{2n}-1$ are simple since $g'(z)=2n z^{2n-1}$ vanishes only in $0$. $\endgroup$ – Jack D'Aurizio Jan 18 '15 at 14:02
  • $\begingroup$ @JackD'Aurizio Can you please show me how this relates to the order of the poles? What result links the derivative of the function in the denominator to the poles? $\endgroup$ – Max Muller Jan 18 '15 at 15:12
  • $\begingroup$ The order of the pole of $\frac{1}{f(z)}$ in $z=z_0$ is just the order of the zero of $f(z)$ in $z=z_0$. $\endgroup$ – Jack D'Aurizio Jan 18 '15 at 15:14

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