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Exercise 1116Algebra › PolynomialsL1Added 3 Nov 2026

Sum of roots of unity

#roots of unity#sum#cyclotomic

Statement

Let ω=e2iπ/n\omega=e^{2i\pi/n}. Show that 1+ω+ω2+⋯+ωn−1=01+\omega+\omega^2+\cdots+\omega^{n-1}=0 for n≥2n\ge2.

Hint

Use the geometric series formula.

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Solution

Geometric series. Since ωn=1\omega^n=1 and ω≠1\omega\neq1 (for n≥2n\ge2),
∑k=0n−1ωk=1−ωn1−ω=1−11−ω=0.\sum_{k=0}^{n-1}\omega^k=\frac{1-\omega^n}{1-\omega}=\frac{1-1}{1-\omega}=0.

Conclusion. The sum of all nn-th roots of unity is zero.

Interpretation. The centroid of the regular nn-gon inscribed in the unit circle is at the origin.