Proving that complex numbers form a regular pentagon Let there be five nonzero complex numbers having the same absolute value and such that zero is equal to their sum, which is equal to the sum of their squares. Prove that the affixes of these numbers in the complex plane form a regular pentagon.
I am familiar with the topic of roots of unity. We have to prove that a,b,c,d,e are roots of $x^5 = 1$. I do not know how to prove starting from $a+b+c+d+e = 0$ and $a^2+b^2+c^2+d^2+e^2=0$ 
 A: Let $|a| = R$. Then $\frac{R^2}{a} = \overline a$, and thus
$$ 0 = \overline a + \overline b + \overline c + \overline d + \overline e = \frac{R^2}{a} + \frac{R^2}{b} + \frac{R^2}{c} + \frac{R^2}{d} + \frac{R^2}{e} = \frac{R^2}{abcde}\left(bcde + acde + abde + abce + abcd\right).$$
Thus, $\sum_{sym}abcd = 0$. Similarly, we obtain
$$ 0 = \overline a^2 + \overline b^2 + \overline c^2 + \overline d^2 + \overline e^2 = \frac{R^4}{a^2} + \frac{R^4}{b^2} + \frac{R^4}{c^2} + \frac{R^4}{d^2} + \frac{R^4}{e^2} = R^4\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} + \frac{1}{e^2}\right).$$
Subtracting the given identites, we infer that
$$0 = \frac{1}{2}\left(\left(a+b+c+d+e\right)^2-(a^2+b^2+c^2+d^2+e^2)\right) = \sum_{sym} ab,$$
and by an anologous trick,
$$ 0 = \frac{1}{2}\left(\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{e}\right)^2-\left(\frac{1}{a} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} + \frac{1}{e^2}\right) \right) = \sum_{sym}\frac{1}{ab} = \frac{1}{abcde}\sum_{sym}cde.$$
Altogether, we now know that $0 = \sum_{sym}a = \sum_{sym}ab = \sum_{sym}abc = \sum_{sym}abcd$. Finally, $abcde = R^5e^{5i \theta }$ for some $\theta \in [0,2\pi)$. Therefore, by Vieta's theorem, $a,b,c,d,e$ are the solutions of
$$ x^5 - R^5e^{5i \theta} = 0,$$
and these are $\{Re^{i\theta + 2\pi i\frac{k}{5}}, k = 1,\dots,5\}$, the vertices of a regular pentagon.
