Fun with 2018 #7 (Roots of the unity #2)

a) $(\omega_{k})^{2019} = 1 \Rightarrow \frac{1}{\omega_k} = (\omega_{k})^{2018}$ , $\forall k \in \mathbb{N}$ such that $1 \leq k \leq 2018$ . By contradiction, if $0 < a < 2018$ satisfies $\frac{1}{\omega_k} = (\omega_{k})^{a}$ $\forall k \in \mathbb{N}$ such that $1 \leq k \leq 2018$ then the equation $x^{a + 1} = 1$ has at least 2019 roots, all $\omega_k$ and $x = 1$ , but $a + 1 < 2018 + 1 = 2019$ and this is a contradiction with fundamental theorem of Algebra.

b) $\forall \space \omega \in \mathbb{C}$ satisfying $P(\omega) = \omega^{2018} + \omega^{2017} + ... + \omega + 1 = 0$ we have $0 = \omega \cdot ( \omega^{2018} + \omega^{2017} + ... + \omega + 1) = \omega^{2019} + (\omega^{2018} + ... + \omega^2 + \omega) = \omega^{2019} - 1$ $\Rightarrow \omega^{2019} = 1 \Rightarrow \omega^{2018} = \frac{1}{\omega}$ . With the same before reasoning, $b = 2018$ is the smallest positive integer satisfying...

Therefore, $\frac{a + b}{2} = \frac{2018 + 2018}{2} = 2018$