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Exercise 838Algebra › Complex numbersL1Added 3 Nov 2026

Sum of the $n$-th roots of unity

#roots of unity#sum#geometric series

Statement

Let ω=e2iπ/n\omega=e^{2i\pi/n}. Prove that
1+ω+ω2+⋯+ωn−1=0.1+\omega+\omega^2+\cdots+\omega^{n-1}=0.

Hint

Use the geometric series formula.

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Solution

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

Conclusion. The sum of all nn-th roots of unity is 00.

Geometric interpretation. The centroid of a regular nn-gon inscribed in the unit circle is the origin.