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When finding the poles of something like the following transfer function, would I be able to write $z=\sqrt[L]{\mu}$ since square roots aren't technically defined on the complex plane?

$$Y(z) = \frac{a_L z^L + a_{L-1}z^{L-1} + ... +a_0}{z^L - \mu}$$

where $a_j,\mu \in\mathbb{C}$, $|\mu|<1$ and L is positive.

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    $\begingroup$ You're mistaken if you think that roots of complex numbers aren't defined: link $\endgroup$ – Matt L. Mar 4 at 14:09
  • $\begingroup$ I tried finding the roots of x^5 - 1 = 0 on Wolfram Alpha There is the obvious solution x = 1 and 4 non-obvious solutions x~~-0.30902 ±0.95106 i x~~0.80902 ± 0.58779 i $\endgroup$ – Ben Mar 4 at 14:09
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    $\begingroup$ So to answer your question, some poles will have imaginary values. $\endgroup$ – Ben Mar 4 at 14:10
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For $\mu > 0$, $z^L \in \left ( \sqrt[L]{\mu} \right) e^{j 2\pi n/L} \,\forall\, n \in 1 \cdots L $.

I.e., the roots are evenly spaced on a circle $\sqrt[L]{\mu}$ in radius, and there's an $L$ of a lot of them.

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    $\begingroup$ You may have made my day with that maths joke haha $\endgroup$ – luke glover Mar 5 at 10:15
  • $\begingroup$ Would we have to consider the case for $\mu \leq 0$? ... well I guess $\mu = 0$ just gives $z=0$ but not sure for $\mu <0$. Thanks for your help :) $\endgroup$ – luke glover Mar 5 at 10:17
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    $\begingroup$ I'm already concerned that I'm answering a homework problem. For $\mu$ equals any number, $z^L - \mu = 0$ is just an algebra problem -- can you work out what the answer is for $\mu < 0$? $\endgroup$ – TimWescott Mar 5 at 14:43
  • $\begingroup$ Not a homework problem, I am researching the design of digital filters but I am new to the subject and its been a while since I formally studied any complex analysis. If I was to make a guess, I would assume that calculating the poles for $\mu < 0$ would just give the same roots of unity again but in reverse direction?(each subsequent root would work clockwise around the unit disc) $\endgroup$ – luke glover Mar 5 at 16:35

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