TITLE: For $a_i \in \mathbb{C}$, does $(1+a_1^k)(1+a_2^k)\cdots (1+a_n^k)=1$ for any positive integer $k$ imply $a_1=\cdots =a_n =0$?
QUESTION [7 upvotes]: For $a_i \in \mathbb{C}$, if $(1+a_1^k)(1+a_2^k)\cdots (1+a_n^k)=1$ holds for all positive integers $k$, does it follow that $a_1=\cdots =a_n =0$?
In fact, I want to prove $|I+A^k|=1$ for any positive integer $k$ ,where $A$ is a $n \times n$ matrix implies $A$ is nilpotent.

REPLY [2 votes]: This is an insane overkill, but it was fun to think about it, so why not to share it?
Let us consider the polynomial $p(z)=\prod_{j=1}^{n}(z-a_j)$ and divide its roots in three groups:

Roots with $|a|<1 $ (group $G_1$ with cardinality $K_1$)
Roots with $|a|=1$ (group $G_2$ with cardinality $K_2$)
Roots with $|a|> 1$ (group $G_3$ with cardinality $K_3$).

My initial purpose is to show that, given the premises, $G_2$ and $G_3$ are actually empty. Since for any $k\in\mathbb{N}$ we have that $\prod_{j=1}^{n}(1+a_j^k)=1\neq 0$ the roots belonging to $G_2$ can only be of the form $\exp\left(2\pi i\frac{m}{d}\right)$ with $d$ odd, or of the form $\exp\left(2\pi i\alpha\right)$ with $\alpha\not\in\mathbb{Q}$. In any case, due to simultaneous diophantine approximation, by carefully choosing $k$ we have that $\prod_{a_j\in G_2}(1+a_j^k)$ can be made arbitrarily close to $2^{K_2}$, while $\prod_{a_j\in G_1}(1+a_j^k)$ converges to $1$ and $\prod_{a_j\in K_3}(1+a_j^k)$ diverges. This proves $K_2=K_3=0$, i.e. all the roots of $p(z)$ lie inside the region $|z|<1$.
Now it comes a clever trick from complex analysis:
$$ \sum_{j=1}^{n}\log(1+a_j^k) = \frac{1}{2\pi i}\oint_{|z|=1}\frac{p'(z)}{p(z)}\phantom{.}\log(1+z^k)\phantom{.} dz $$
where everything is well-defined and the RHS equals $0$ for any $k\in\mathbb{N}$. For a suitable arithmetic function $c_n$ (computable via Moebius inversion) we have
$$ z = \sum_{n\geq 1} c_n \log(1+z^n) $$
hence the previous point implies
$$ \sum_{j=1}^{n} a_j^k = \frac{1}{2\pi i}\oint_{|z|=1}\frac{p'(z)}{p(z)}z^k\phantom{.}dz = 0$$
and all the power sums are zero. Newton's formulas hence imply that all the elementary symmetric functions of $a_1,\ldots,a_n$ are zero, so $p(z)=z^n$ and $a_1=a_2=\ldots=a_n=0$ as wanted.