Easy but complex

Algebra Level pending

A complex number $z$ follows the following formula

$\sqrt[2n+1]{z} = 1, r_1, \ldots r_{2n+1}$

Where $n$ is any positive integer which satisfies

$n \geq 1$

and $r_x$ is a root.

Out of the following, which is equal to the sum of the roots.

0 -z 1 z

1 solution

Jack Rawlin
Jan 5, 2015

Since a root of $z$ is $1$ that means that $z$ has $2n+1$ roots of unity thus the following equation holds true

$1 + r_1 + r_2 + \ldots + r_{2n + 1} = 0$