Find all nonconstant polynomials $P(z)$ with complex coefficients for which all complex roots of the polynomials $P(z)$ and $P(z)-1$ have absolute value 1.
The answer is $P(x)$ should be a polynomial of the form $P(x)=\lambda x^{n}-\mu$ where $|\lambda|=|\mu|$ and $\operatorname{Re} \mu=-\frac{1}{2}$. One may check these all work; let's prove they are the only solutions. \l First approach (Evan Chen). We introduce the following notations:  $$ \begin{aligned} P(x) & =c_{n} x^{n}+c_{n-1} x^{n-1}+\cdots+c_{1} x+c_{0} \\ & =c_{n}\left(x+\alpha_{1}\right) \ldots\left(x+\alpha_{n}\right) \\ P(x)-1 & =c_{n}\left(x+\beta_{1}\right) \ldots\left(x+\beta_{n}\right) \end{aligned} $$  By taking conjugates,  $$ \begin{aligned} \left(x+\alpha_{1}\right) \cdots\left(x+\alpha_{n}\right) & =\left(x+\beta_{1}\right) \cdots\left(x+\beta_{n}\right)+c_{n}^{-1} \\ \Longrightarrow\left(x+\frac{1}{\alpha_{1}}\right) \cdots\left(x+\frac{1}{\alpha_{n}}\right) & =\left(x+\frac{1}{\beta_{1}}\right) \cdots\left(x+\frac{1}{\beta_{n}}\right)+\left(\overline{c_{n}}\right)^{-1} \end{aligned} $$  The equation $(\boldsymbol{\oplus})$ is the main player: Claim - We have $c_{k}=0$ for all $k=1, \ldots, n-1$.   $$ \frac{c_{n-k}}{\prod_{i} \alpha_{i}}=\frac{c_{n-k}}{\prod_{i} \beta_{i}} $$  but $\prod_{i} \alpha_{i}-\prod_{i} \beta_{i}=\frac{1}{c_{n}} \neq 0$. Hence $c_{k}=0$. It follows that $P(x)$ must be of the form $P(x)=\lambda x^{n}-\mu$, so that $P(x)=\lambda x^{n}-(\mu+1)$. This requires $|\mu|=|\mu+1|=|\lambda|$ which is equivalent to the stated part.