Consider nth roots of unity, show that the sum of their pth powers always vanishes unless p is a multiple of n, p being an integer, and then show that the sum is n.

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Solution

$n t h$ roots of unity are $1, a, a^{2}, . . ., a^{n - 1}$
where $a = cos (\frac{2 π}{n}) + i sin (\frac{2 π}{n})$
Therefore the sum of the $p t h$ power of these roots $= 1^{p} + a^{p} + a^{2 p} + a^{3 p} + . . . a^{(n - 1) p}$ ...(1)
Case 1) If $p$ is not a multiple of $n$ , we have
$a^{p} = {[cos (\frac{2 π}{n}) + i sin (\frac{2 π}{n})]}^{p} = cos [2 π (\frac{p}{n})] + i sin [2 π (\frac{p}{n})] \neq 1$
Because $(\frac{p}{n})$ is not an integer. So in this case summing the G.P in (1) whose common ratio $a^{p}$ in not 1, we have
the sum of the $p t h$ powers of the roots
$= \frac{{1 - (a^{p})}^{n}}{1 - a^{p}} = \frac{1 - a^{p n}}{1 - a^{p}} = \frac{1 - {(a^{n})}^{p}}{1 - a^{p}} = \frac{1 - 1}{1 - a^{p}}$
Since $a^{n} = 1, a$ being $n t h$ root of unity
$= \frac{0}{1 - a^{p}} = 0, a^{p} \neq 1$
Case 2) If $p$ is a multiple of $n$ , say $p = m n$ , where $m$ is integer, then $a^{p} = a^{m n} = {(a^{n})}^{m} = 1^{m} = 1$
So in this case each term in (1) is equal to 1 and the sum of the $p t h$ powers of the roots
$= 1 + 1 + 1 + . . .1$ (upto n terms ) $= n$

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