# Sum of the Nth Roots of Unity

Let $\epsilon _{0} = 1, \ldots , \epsilon _{N-1}$ be the vertices of a regular $N$-gon inscribed on the unit circle. Show that the sum of all $\epsilon _{k}, k = 0, \ldots , N-1,$ equals zero.

After a suitable adjustment (rotation) of the axes, the vertices of a regular $N$-gon inscribed in a unit circle can be identified with the $N^{th}$ roots of unity, so that $\epsilon _{k} = e^{2\pi k\cdot i/N}.$ Thus we are looking into the value of the sum

$\displaystyle S = \sum_{k=0}^{N-1} \epsilon _{k} = \sum_{k=0}^{N-1} e^{2\pi k\cdot i/N},$

In [Trigg, #213] the problem is posed a little differently: Prove that the sum of all vectors from the center of a regular $N$-gon to its vertices is zero.

### Reference

- C. W. Trigg,
*Mathematical Quickies*, Dover, 1985

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Copyright © 1996-2018 Alexander Bogomolny

Let $\epsilon _{0} = 1, \ldots , \epsilon _{N-1}$ be the vertices of a regular $N$-gon inscribed on the unit circle. Show that the sum of all $\epsilon _{k}, k = 0, \ldots , N-1,$ equals zero.

### Solution

The geometric interpretation is decidedly useful because of its explicit symmetry. A rotation through an angle $2\pi /N$ maps the $N$-gon onto itself. In the realm of complex numbers a rotation through angle $\alpha$ around the origin is enacted by multiplying by $e^{i\alpha }= \cos\alpha + i\cdot \sin\alpha.$ The sum $S$ of the vertices of a regular $N$-gon does not change when multiplied by $e^{2\pi i/N}:$

$S = e^{2\pi i/N}S,$

implying that the sum is zero.

The proof is also easy algebraically:

$\displaystyle \begin{align} e^{2\pi i/N}\sum_{k=0}^{N-1}e^{2\pi i\cdot k/N} &= \sum_{k=1}^{N}e^{2\pi i\cdot k/N}\\ &= \sum_{k=1}^{N-1}e^{2\pi i\cdot k/N} + e^{2\pi i\cdot N/N}\\ &= \sum_{k=1}^{N-1}e^{2\pi i\cdot k/N} + 1\\ &= \sum_{k=1}^{N-1}e^{2\pi i\cdot k/N} + e^{2\pi i\cdot 0/N}\\ &= \sum_{k=0}^{N-1}e^{2\pi i\cdot k/N} \end{align}$

Equating to zero separately the real and the imaginary parts, one obtains two well known trigonometric identities:

$\displaystyle\sum_{k=0}^{N-1} \cos (2\pi i\cdot k/N) = 0,$

$\displaystyle\sum_{k=0}^{N-1} \sin(2\pi i\cdot k/N) = 0.$

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Copyright © 1996-2018 Alexander Bogomolny

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