Exercise
Let $w, z$ are the unit roots of equation $z^5 = 1$ for $z \in \mathbb{C}$.
Prove that $(w^{21} + z^{14})^5$ is always a real number.
So, first I can do:
$w^{21} = w$
$z^{14} = z^4$
So then I have:
$(w + z^{4})^5$
But I'm not sure what I can do with this. I've tried using Newton's Binomial and also trying to write the roots as $e^{i(\frac{\phi + 2k\pi}{5})}$ but because the roots are not multiplied but in a sum I don't think that will be useful. Using Newton's Binomial I ended with a summatory but with different coefficient, so I wasn't able to apply the typical rule to solve these kind of exercises:
If w belongs to Gn:
$\sum_{i = 0}^{n-1} w^{i} = 0$
You're off to a good start. Note that in your simplified expression $(w+z^4)^5$, $w$ and $z^4$ could be any fifth roots of unity. So the problem reduces to showing that the sum of two fifth roots of unity raised to the fifth power yields a real number.
So let $a$ and $b$ be any fifth roots of unity. Note that $\bar{a}=a^4$ and $\bar{b}=b^4$.
We have $(a+b)^5= a^5 + 5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$
Now note that $a^5=1$ and $b^5=1$. So these terms contribute only real values to the sum. But also $a^4b$ and $ab^4$ are conjugates. So $5a^4b+5ab^4$ is real. Similarly, $a^3b^2$ and $a^2b^3$ are conjugates. So $10a^3b^2+10a^2b^3$ is also real.