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I've been learning about roots of unity and how they manifest on the complex plane. I understand that if you take $z^{n}=1$, then the values of $z$ that satisfy this equation happen to lie on the unit circle with equal angles $\frac{2\pi}{n}$ between them.

I tried a couple examples on Wolfram Alpha, here's z^5 = 1 and z^12 = 1. But I was wondering if a similar strategy could be used for arbitrary complex polynomials, so I decided to tweak the inputs to have more terms. For example, here's z^5 + z^3 = 1 and z^5 - z^3 + z = 1.

The complex plane representations generated by Wolfram Alpha seem to suggest that the solutions might lie on an ellipse instead of a pure circle, or maybe some other characteristic curve. Is this the case?

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  • $\begingroup$ At least for a quintic, the points lie on a conic, although not necesseraly an elipse: it can be a circle, parabolla or even hyperbola. I do not know where there is some formula for higher degree polynomials, good question. $\endgroup$
    – RicardoMM
    Apr 15 '21 at 10:04
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    $\begingroup$ Not in general. But see the nice pictures in math.stackexchange.com/questions/535720/…, especially math.stackexchange.com/a/109605/589 $\endgroup$
    – lhf
    Apr 15 '21 at 10:10
  • $\begingroup$ @RicardoMM I see, interesting that it's known for quintics specifically but not necessarily for other polynomials. Can you recommend me any resources to learn more about complex analysis (granted I get more familiar with complex numbers first)? $\endgroup$ Apr 15 '21 at 10:12
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    $\begingroup$ @JansthcirlU I think Conway's book is a nice textbook books.google.ca/books?id=9LtfZr1snG0C&hl=pt-BR. Of course it depends on your field of study. $\endgroup$
    – RicardoMM
    Apr 15 '21 at 10:30
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    $\begingroup$ @JansthcirlU In special cases there may be a sort of symmetry, but I do not know under what conditions we have that. See the comment to my answer. $\endgroup$
    – Paul Frost
    Apr 16 '21 at 22:14
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Of course you can always find a closed polygonal chain containing all roots, and you can also achieve that it has no self-intersections. If you want, you can also smooth this curve which will give you a smooth Jordan curve.

But you cannot expect that this is a particular nice curve. In fact, any finite set of complex numbers is the set of roots of some complex polynomial. Such a finite set does in general not lie on an ellipse or any other "regular" curve.

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  • $\begingroup$ So if I'm interpreting your answer correctly, you're saying that it's possible to construct a curve that includes the points, but that that curve isn't necessarily nice, right? But how would fitting the points to a custom made curve help me find that the points do (or do not) lie on a special characteristic function that does happen to be nice? $\endgroup$ Apr 15 '21 at 10:29
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    $\begingroup$ @JansthcirlU There are certainly special cases where you can find a "nice" curve. So you could ask a new question: Can one identify complex polynomials with this "niceness property"? But you should explain, at least on an intuitive level, what nice means. For general polynomials you do not have a chance because then you essentially consider all finite sets of complex numbers. $\endgroup$
    – Paul Frost
    Apr 15 '21 at 11:06

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