## Task A-4.1.

Determine all complex numbers $a$ for which all coefficients of the polynomial

$$
P(x)=(x-a)\left(x-a^{2}\right)\left(x-a^{3}\right)\left(x-a^{4}\right)
$$

are real.
## First Solution.

Every real number $a$ satisfies the required condition, so let's assume further that $a$ is not a real number.

Let $z$ be a complex number that is not real. If $z$ is a root of a polynomial with real coefficients, then $\bar{z}$ is also a root. Therefore, we have one of three cases: $\bar{a}=a^{2}$, $\bar{a}=a^{3}$, or $\bar{a}=a^{4}$.

If $\bar{a}=a^{2}$, then $|a|=|a|^{2}$, and since $a \neq 0$, it follows that $|a|=1$. Furthermore, we get $a^{3}=a^{2} \cdot a=\bar{a} \cdot a=|a|^{2}=1$. Therefore, $a^{3}$ is a real root of the polynomial $P(x)$, and so must be $a^{4}$, since it cannot form a complex conjugate pair with any other root. Therefore, $a=\frac{a^{4}}{a^{3}}$ is a real number, which is a contradiction.

If $\bar{a}=a^{3}$, similarly, we show that $a^{4}=1$. From here, $a= \pm i$, since we are not considering real solutions. By direct verification, we see that both possibilities for $a$ satisfy the conditions of the problem

$$
P(x)=(x \mp i)(x+1)(x \pm i)(x-1)=\left(x^{2}+1\right)\left(x^{2}-1\right)=x^{4}-1 .
$$

If $\bar{a}=a^{4}$, similarly, we show that $a^{5}=1$. The solutions that are not real are

$$
a=\cos \left(\frac{2 k \pi}{5}\right)+i \sin \left(\frac{2 k \pi}{5}\right), k=1,2,3,4
$$

Notice that from the fact that $a^{5}=1$, it follows that

$$
\bar{a}^{2}=a^{5} \bar{a}^{2}=a^{3} a^{2} \bar{a}^{2}=a^{3}|a|^{4}=a^{3}
$$

Thus, for every complex number $a$, which is not real, such that $a^{5}=1$, the numbers $a^{2}$ and $a^{3}$ also form a complex conjugate pair. Since the pairs $\left(a, a^{4}\right)$ and $\left(a^{2}, a^{3}\right)$ are complex conjugates, we conclude that the polynomial $P(x)$ has real coefficients.

Finally, all complex numbers that satisfy the required condition are all real numbers $a$ and

$$
a \in\{i,-i\} \cup\left\{\cos \left(\frac{2 k \pi}{5}\right)+i \sin \left(\frac{2 k \pi}{5}\right): k=1,2,3,4\right\}
$$