If the roots of unity lie on a circle, do arbitrary polynomial roots also lie on some kind of characteristic curve?

Question

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?

Answer 1

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.