sum of the roots of unity

Question

$$\sum_{k=0}^{N-1} e^{-j\frac{2\pi}{N}(i + n)k} = \begin{cases} N & \text{for } (i + n) = 0,N,2N,3N, \ldots \\ 0 & \text{otherwise} \end{cases} = N\delta[(i + n)\text{ mod }N]$$

Can someone provide an explanation why the sum is equal to zero if (i+n) mod N is not equal to zero?

You have been around for almost three years. Haven't you yet noticed that you are supposed to use MathJax around here? — José Carlos Santos, Feb 14 '21 at 21:34

score 1 · Answer 1 · answered Feb 14 '21 at 22:13

1

For any polynomial where the coefficient of the leading term $=1$, then the coefficient of the second term is the negative sum of the roots. Therefore for $z^n-1=0$, the sum$=0$.

answered Feb 14 '21 at 22:13

herb steinberg

12,765

for $n>1$, that is – J. W. Tanner Feb 14 '21 at 22:18
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@J.W.Tanner You are right, although the question was presented so that it didn't matter. – herb steinberg Feb 16 '21 at 04:21

sum of the roots of unity

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