Suppose I'm given a polynomial $P(z)=z^n + a_{n-1}z^{n-1}+\ldots+a_0$ in closed unit disk and know that $|P(z)|$ doesn't exceed $1$ in the domain: then

$P(z) = z^n$ is it's only possible form

My work.
If $P(z)=z^n+a_{n-1}z^{n-1}+\ldots+a_0$ then $$P(1)=1+a_{n-1}+\ldots+a_0$$ and since $|P(1)| \le 1$ we have $|1+a_{n-1}+\ldots+a-0|$$\le 1$.

I'm just stuck here not able to proceed any further Any idea as how to proceed.

  • $\begingroup$ You need to require that modulus 1 be achieved on the closed disc, otherwise any fixed polynomial multiplied by a small enough positive real would work. Not sure what else might be required... $\endgroup$ – Will Jagy Oct 20 '20 at 4:15
  • $\begingroup$ @WillJagy Did you miss that the coefficient of $z^n$ is given as $1$? $\endgroup$ – Robert Israel Oct 20 '20 at 4:29
  • $\begingroup$ @RobertIsrael I did miss it, good catch $\endgroup$ – Will Jagy Oct 20 '20 at 14:15

Define $a_n = 1$, and $a_k= 0$ for $k \notin [0,1,\ldots,n]$. The $a_k$ are the Fourier coefficients of $P(e^{i\theta})$ as a function on $[0,2\pi]$. Parseval's theorem says $$\sum_{k=0}^n |a_k|^2 = \frac{1}{2\pi} \int_0^{2\pi} |P(e^{i\theta})|^2 \; d\theta $$ Since $a_n = 1$, the left side $\ge 1$. But since $|P(e^{i\theta})|\le 1$, the right side $\le 1$. Therefore both sides are $1$, and so $|a_k|^2 = 0$ for $0 \le k \le n-1$.


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