Can a nonconstant polynomial in $\mathbb{C[x]}$ take only real values on the unit circle?

Question

Conjecture: If a polynomial in $\mathbb{C[x]}$ take only real values on the unit circle, then its identically equal to a real constant.

It's quite easy to see that the statement can be reformulated as:

Let $r_1,...,r_n$ and $\phi_1,...\phi_n$ be real numbers. If $S(\alpha)=\sum\limits_{k=1}^{n}r_k\sin(\phi_k+k\alpha)$ is constant for all $\alpha \in \mathbb{R}$, then $r_1=r_2=...=r_n=0$.

This seems intuitively true, because its essentially saying that we can't have a sum of harmonics which sum to zero.

Can I show the reformulation via simples means without resorting to Fourier series?

score 1 · Answer 1 · answered Jan 23 '17 at 18:13

1

You can just use the maximum principle for harmonic functions. If any function analytic in the open unit disc and continuous on the boundary is purely real on the boundary, then by the maximum principle applied to its imaginary part it is purely real on the entire unit disc, and so is constant.

answered Jan 23 '17 at 18:13

Vik78

3,877

is there a more elementary way at arriving at the answer which doesn't use complex analysis? – math_lover Jan 23 '17 at 18:15
Not that I know of. Maybe there's a Fourier approach, but I don't know enough to see it. T – Vik78 Jan 23 '17 at 18:20

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

The functions $e_k:\mathbb R\to\mathbb C$ defined by $e_k(\theta)=e^{ik\theta}$ are linearly independent for any set of distinct $k$, in particular for $k\in\mathbb Z$. One way to see this is because eigenvectors for distinct eigenvalues of a linear transformation on a vector space are linearly independent, applied on the complex vector space $V$ of infinitely differentiable functions from $\mathbb R$ to $\mathbb C$ with the linear transformation being differentiation, $D:V\to V$, $Df = f'$. Note that $De_k=ike_k$, so $e_k$ is an eigenvector for $D$ with eigenvalue $ik$.

Suppose $f\in \mathbb C[x]$, so $f(x)=\sum\limits_{k=0}^n a_k x^k$. On the unit circle, $f(e^{i\theta})=\sum\limits_{k=0}^n a_k e^{ik\theta}$. If this quantity is always real, then $a_0-\overline{a_0}+\sum\limits_{k=1}^{n}a_k e^{ik\theta}+\sum\limits_{k=-n}^{-1}-\overline{a_{-k}}e^{ik\theta}=0$ for all $\theta\in\mathbb R$. In other words, $$(a_0-\overline{a_0})e_0+\sum\limits_{k=1}^{n}a_k e_k+\sum\limits_{k=-n}^{-1}-\overline{a_{-k}}e_k=0.$$

Because $\{e_k: k\in\{-n,-n+1,\ldots,n-1,n\}\}$ is linearly independent, each coefficient must be zero, so $a_0$ is real and $a_k=0$ for $k>0$.

Very beautiful. I just started learning about vector spaces and I didn't know this result, but this is clearly key. +1 for providing an elementary solution — math_lover, Jan 25 '17 at 22:21

Can a nonconstant polynomial in $\mathbb{C[x]}$ take only real values on the unit circle?

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