## Task A-4.1.

Let $n$ be a natural number. Determine all complex numbers $z$ such that

$$
\left(1-z+z^{2}\right)\left(1-z^{2}+z^{4}\right)\left(1-z^{4}+z^{8}\right) \cdots\left(1-z^{2^{n-1}}+z^{2^{n}}\right)=\frac{3 z^{2^{n}}}{1+z+z^{2}}
$$
## Solution.

First, it is necessary that $1+z+z^{2} \neq 0$, which means that $z$ must not be such that $z^{3}=1$ and $z \neq 1$. That is, $z$ is not one of the two cube roots of 1 different from 1.

Multiply the entire equation by $1+z+z^{2}$. We get the equation

$$
\left(1+z+z^{2}\right)\left(1-z+z^{2}\right)\left(1-z^{2}+z^{4}\right)\left(1-z^{4}+z^{8}\right) \cdots\left(1-z^{2^{n-1}}+z^{2^{n}}\right)=3 z^{2^{n}}
$$

Let $I_{n}$ denote the left side of the last equality. We will prove by mathematical induction on $n$ that $I_{n}=1+z^{2^{n}}+z^{2^{n+1}}$. In the proof, we will use the auxiliary claim: for every natural number $k$,

$$
\left(1+z^{k}+z^{2 k}\right)\left(1-z^{k}+z^{2 k}\right)=\left(1+z^{2 k}\right)^{2}-\left(z^{k}\right)^{2}=1+z^{2 k}+z^{4 k}
$$

Let's prove the claim. For the base case, $n=1$, we have

$$
I_{1}=\left(1+z+z^{2}\right)\left(1-z+z^{2}\right)=1+z^{2}+z^{4}
$$

where we used the auxiliary claim for $k=1$. Assume the induction hypothesis holds for some $n$, and prove it for $n+1$:

$$
\begin{aligned}
I_{n+1} & =I_{n} \cdot\left(1-z^{2^{n}}+z^{2^{n+1}}\right) \\
& =\left(1+z^{2^{n}}+z^{2^{n+1}}\right)\left(1-z^{2^{n}}+z^{2^{n+1}}\right) \\
& =1+z^{2^{n+1}}+z^{2^{n+2}}
\end{aligned}
$$

where we used the induction hypothesis in the second-to-last equality and the auxiliary claim for $k=2^{n}$ in the last equality. This completes the induction proof.

Thus, we have the equation

$$
1+z^{2^{n}}+z^{2^{n+1}}=3 z^{2^{n}}
$$

or

$$
z^{2^{n+1}}-2 z^{2^{n}}+1=0, \quad \text { i.e. } \quad\left(z^{2^{n}}-1\right)^{2}=0
$$

Finally, the solutions are all the $2^{n}$-th roots of 1, i.e.,

$$
z_{k}=\cos \frac{k \pi}{2^{n-1}}+i \sin \frac{k \pi}{2^{n-1}}, \quad \text { for } k \in\left\{0,1,2, \ldots, 2^{n}-1\right\}
$$

All these numbers are indeed solutions, as no $2^{n}$-th root of 1 is also a cube root of 1, except for the number 1 itself.