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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?

JansthcirlU
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  • 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. – RicardoMM Apr 15 '21 at 10:04
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    Not in general. But see the nice pictures in https://math.stackexchange.com/questions/535720/what-are-the-properties-of-the-roots-of-the-incomplete-finite-exponential-series, especially https://math.stackexchange.com/a/109605/589 – lhf Apr 15 '21 at 10:10
  • @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)? – JansthcirlU Apr 15 '21 at 10:12
  • @lhf that's a great link, I'll definitely be looking into the resources mentioned once I get up to speed, thanks! – JansthcirlU Apr 15 '21 at 10:14
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    @JansthcirlU I think Conway's book is a nice textbook https://books.google.ca/books?id=9LtfZr1snG0C&hl=pt-BR. Of course it depends on your field of study. – RicardoMM Apr 15 '21 at 10:30
  • @RicardoMM On which conic do $0,1,2,3, i$ lie? – Paul Frost Apr 15 '21 at 11:23
  • Not a curve, but if your polynomial has real coefficients, the roots lie symmetric around the real line (so if there is a nice curve, it is mirrored at that axis). The circle for the roots of unity is actually a special case of this. – Torsten Schoeneberg Apr 15 '21 at 13:53
  • As the answer points out, as long as you have finitely many points, you can put plenty of pretty but useless curves through them which will not tell you much. A better question starting from the unit roots might be: Is there another (countable) family of polynomials $p_n$, parametrized somewhat nicely, such that the roots of all $p_n$ lie densely on some smooth curve? – Torsten Schoeneberg Apr 15 '21 at 14:05
  • Also, if anyone writes an answer containing the words "elliptic curve" "complex multiplication" and "class field theory" or "Kronecker-Weber", I'll upvote it. If it also contains "Lubin-Tate", I will give it a bonus. – Torsten Schoeneberg Apr 15 '21 at 14:06
  • Look at https://www.wolframalpha.com/input/?i=z%5E5+-+3z%5E3+-5z%5E2%2B4z+%3D+1. Note that if your polynomial has real coefficients, then the set of roots is symmetric with respect to the real axis. See also https://www.wolframalpha.com/input/?i=z%5E5+-+iz%5E3+-5iz%5E2%2B4z+%3D+1. – Paul Frost Apr 15 '21 at 23:29
  • @PaulFrost it looks like that second plot has an axis of symmetry about the argument of $-1.4431+0.9696i$, is that a coincidence? Well, not a symmetry but it seems that the arguments of the other roots are mirrored about it. – JansthcirlU Apr 16 '21 at 14:10
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    @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. – 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.

Paul Frost
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  • 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? – JansthcirlU Apr 15 '21 at 10:29
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    @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. – Paul Frost Apr 15 '21 at 11:06