Let $p$ be a polynomial of degree $n$ such that $|p(z)| = 1$ for all $|z| = 1$.

Why is it that $p(z) = az^n$ for some $|a| = 1$?

I've noticed that we could easily prove this by induction if we could show that 0 was a root of $p$. My guess is that Rouche's theorem and/or the Maximum Modulus principle should be used.

My background to the problem: This question is the last question on the take-home portion of my final exam that I haven't been able to figure out yet.

We are allowed to collaborate with other people in the class (and I have) as well as use any book and the internet (including this site). So basically, it's a homework assignment that is worth more points than usual. Nevertheless, in case it helps you decide how much information to give, the final is due tomorrow (but since I'm going to a math conference, I may turn it in late).


3 Answers 3


If $p$ has no zeros in the unit disk, then $p$ must be constant - consider $\frac{1}{p}$.

So for $n > 0$, $p$ must have zeros in the unit disk. Let them be $0 = \zeta_0, \zeta_1,\dotsc,\zeta_k$, with multiplicities $\mu_0,\dotsc,\mu_k$, where we allow $\mu_0 = 0$, but demand $\mu_\kappa > 0$ for $1 \leqslant \kappa \leqslant k$. Consider

$$f(z) = z^{\mu_0} \prod_{\kappa = 1}^k \left(\frac{z-\zeta_\kappa}{1-\overline{\zeta_\kappa}\cdot z}\right)^{\mu_\kappa}.$$

$f$ is holomorphic in a neighbourhood of the closed unit disk, and $f$ has the same zeros (including multiplicity) in the unit disk as $p$. Hence

$$g(z) = \frac{f(z)}{p(z)}$$

is holomorphic in a neighbourhood of the closed unit disk and has no zeros in the unit disk. For $\lvert z\rvert = 1$, you can say something about $g(z)$ that implies that $g$ is constant. And that implies that $k = 0$ and $\mu_0 = n$.

Alternatively, consider the entire meromorphic function

$$h(z) = \frac{1}{\overline{p(1/\overline{z})}}.$$

Use the conditions on $p$ to deduce $p = h$, and from that the result.

  • $\begingroup$ Thank you. Small correction on the first solution: yes we want $f$ to have all the same zeros including multiplicities as $p$, but as it is now, for $j > 0$, the multiplicity of the root $\zeta_j$ is 1. $\endgroup$ May 9, 2014 at 0:31
  • $\begingroup$ Aw, crap, I forgot the powers. Thanks for the heads-up. $\endgroup$ May 9, 2014 at 0:36

...some comments that don't necessarily go anywhere (because I need to go).

  • define $p(e^{i\theta})=e^{iP(\theta)}$ with $P:\mathbb{R}\rightarrow \mathbb{R}$, $2\pi$-periodic and of course $p((e^{i\theta})^n=p(e^{in\theta}))=e^{iP(n(\theta))}$
  • let $p(e^{i\theta})=\sum_{\ell=0}^n\alpha_\ell e^{i\ell\theta}$ and this gives $\left|\sum_{\ell=0}^n\alpha_\ell\right|=1$
  • $\displaystyle \frac{d^np}{dz^n}=n!\alpha_n$ and $\displaystyle \frac{d^np}{d\theta^n}=\sum_{\ell=0}^n\alpha_\ell (i\ell)^ne^{i\ell \theta}$
  • could we show $p(\zeta_n^\ell)=\alpha_n$ for $\zeta_n$ a primitive $n$-th root of unity and $k=1,2,\dots,n$ and $p(0)=0$ using $\sum_{\ell=0}^n\zeta_n^\ell=0$? I think we can show that for all $\pi\in S_{n-1}$ that $$\left|\alpha_n+\alpha_0+\sum_{\ell=1}^{n-1}\alpha_\ell \zeta_n^{\pi(\ell)}\right|=1.$$

This $$\sum_{n=0}^{\infty}|a_nz^n|^2=\frac{1}{2 \pi}\int_{-\pi}^{\pi}|p(ze^{it})|^2dt$$ kills it if you suppose that the leading coefficient of $p$ has modulus $1$.

I'm working on the general case.


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