I know the cyclotomic polynomials can be defined as such (from Wikipedia):

The monic polynomials with integer coefficients that are the minimal polynomial over the field of the rational numbers of any primitive nth-root of unity.

And that they can be described and generated algorithmically.

But what if we take away the "integer coefficient" restriction? I know polynomials like $(x+i)(x+1)(x-1)$ have roots of unity as their roots. So, I have three questions:

  1. If we allow real coefficients, do we get any more polynomials with roots of unity roots?

  2. Is there a general pattern behind or way of generating all complex polynomials with roots of unity as roots?

  • $\begingroup$ $\bf{all}$ nth-roots. In your example, the polynomial must be $(x+1)(x-1)(x+i)(x-i)$ that is the polynomial with $\bf{all}$ solutions of $z^4=1$ $\endgroup$ Jan 9, 2018 at 1:16
  • $\begingroup$ Do you require that they be roots of unity, or just numbers on the unit circle? I believe the second one has a nicer characterization, at least with real coefficients. $\endgroup$ Jan 9, 2018 at 1:16
  • $\begingroup$ @CarlSchildkraut Specifically roots of unity, sadly, I can't settle here. $\endgroup$
    – Nico A
    Jan 9, 2018 at 1:20
  • $\begingroup$ The answer to 1 is definitely yes. If we allow real coefficients, such polynomials are generated (multiplicatively) by $(x-1)$, $(x+1)$, and $(x-\omega)(x-\bar{\omega})$, where $\omega$ is a root of unity. $\endgroup$
    – jgon
    Jan 9, 2018 at 2:18
  • $\begingroup$ @jgon I believe that also answers 2, if you take it over all possible roots of unity. $\endgroup$ Jan 9, 2018 at 2:22

1 Answer 1


Part 1 is still kind of nice; depending on what you need this for this may or may not be useful:

For a real polynomial, its roots are either real or form conjugate pairs. If you have two conjugate roots of unity

$$e^{\frac{2i\pi k}{n}},\ e^{-\frac{2i\pi k}{n}},$$

then the quadratic polynomial with them as roots is

$$P(x)=x^2-2\cos\left(\frac{2\pi k}{n}\right)x+1.$$

So, any real-coefficiented polynomial (whose roots are all roots of unity) has the form

$$(x+1)^m(x-1)^n\prod_{k=1}^N \left(x^2-2\cos\left(\frac{2\pi a_k}{b_k}\right)x+1\right),$$

where $m,n$ are nonnegative integers and all $a_k,b_k$ are positive integers (with $a_k<b_k$, if you wish).

For part 2, I mean, you can just pick the roots to be

$$x_k=e^{\frac{2i\pi a_k}{b_k}},$$

and your polynomial will be

$$\prod_{k=1}^N \left(x-e^{\frac{2i\pi a_k}{b_k}}\right),$$

but this probably isn't that useful. Another thing that may or may not be useful for either of these is that all polynomials with only roots of unity as roots have roots that all satisfy $x^M=1$ for some (possibly large with respect to the degree of the polynomial) integer $M$. In the case above this $M$ can be taken to be $b_1b_2\cdots b_k$. So, all polynomials with only roots of unity as roots are factors of $x^M-1$ for some $M$.

  • $\begingroup$ Wow, exam week must be having more of an effect on me than I thought because yes, of course - the polynomials with roots of the roots of unity are... the polynomials with factors of (x - the roots of unity). $\endgroup$
    – Nico A
    Jan 9, 2018 at 2:26
  • $\begingroup$ But thank you very much! Very nice and clear answer. $\endgroup$
    – Nico A
    Jan 9, 2018 at 2:26
  • $\begingroup$ @TreFox Thank you. Glad I could help! $\endgroup$ Jan 9, 2018 at 2:26

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