# Complex polynomial in closed unit disk with its mod value not exceeding $1$ is only of the form $z^n$ [duplicate]

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.

• 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... – Will Jagy Oct 20 '20 at 4:15
• @WillJagy Did you miss that the coefficient of $z^n$ is given as $1$? – Robert Israel Oct 20 '20 at 4:29
• @RobertIsrael I did miss it, good catch – Will Jagy Oct 20 '20 at 14:15

## 1 Answer

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 $$0 \le k \le n-1$$.