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

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.

Answer 1

Expand the product $(1+a_1)\cdots (1+a_n)$, we obtain $$ e_1 +\cdots + e_n = 0 $$ where $e_m=\sum_{1\leq j_1<\cdots <j_m\leq n} a_{j_1}\cdots a_{j_m}$ for each $m\leq n$.

Let $e_{m,k}=\sum_{1\leq j_1<\cdots <j_m\leq n} a_{j_1}^k\cdots a_{j_m}^k$.

By the assumption, we obtain $$ e_{1,k}+\cdots + e_{n,k}=0 $$ for each $k\geq 1$.

This condition implies each summand $a_{j_1}\cdots a_{j_m}$ appearing in $e_m$ is zero. For the proof, see this answer by Sangchul Lee.

Since each $a_j$, $1\leq j\leq n$ appears in $e_1$, we have $$ a_1=\cdots =a_n=0 $$ as desired.

Answer 2

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:

1. Roots with $|a|<1 $ (group $G_1$ with cardinality $K_1$)
2. Roots with $|a|=1$ (group $G_2$ with cardinality $K_2$)
3. 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.