### Proposition 1.2.18: Roots of Unity

 The n n-th roots of unity are given by wnk, where k = 0, 1, 2, ... n-1 and wn = cis(2/n) They form the vertices of a regular polygon and add up to zero, i.e. they satisfy the equation: 1 + wn + wn2 + ... + wnn-1 = 0 Context

Our proofs so far, for the most part, were mostly computational and did not require great ingenuity. But the proof of the second part of this statement is different, it needs some smart trickery. The first part, though, is straight foward. Let

wn = cis(2/n)

Then

wnk = cis(2 k / n)

by DeMoivre's formula. Therefore, using that formula again:

(wnk)n = cis(2 k n / n) = cis(2 k ) = 1

for all positive integers k. Thus, each wnk is an n-th root of unity (because its n-th power is 1). And if k = 0, 1, 2, ..., n-1 then all wnk are distinct so that we have indeed found n different n-th roots of 1.

As for the second part, let

Sn = 1 + z + z2 + z3 + ... + zn

Then

z Sn = z + z2 + z3 + ... + zn + zn+1

Subtracting terms we get

Sn - z Sn = Sn(1-z) = 1 - zn+1

because all terms of Sn - z Sn except the first and last remain. But then, for z 1

Sn = 1 + z + z2 + z3 + ... + zn = 1-zn+1/1-z

or equivalently Sn-1 = 1-zn/1-z. But then

(1-z) Sn-1 = (1-z)(1 + z + z2 + z3 + ... + zn-1) = 1 - zn

If we now substitute wn = cis(2/n) for n > 0 we have

(1-wn)(1 + wn + wn2 + ... + wnn-1) = 1 - (wn)n = 0

because wnn = 1. But now our statement follows because wn 1 if n > 0. Now that's a proof.

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Here are, for example, the eight roots of z8=1:

 The 8 roots of z8=1 Sum of 8-th roots is zero

Incidentally, the fact that

(1-z)(1 + z + z2 + z3 + ... + zn-1) = 1 - zn

can also be confirmed by "long division" - but who wants to do that ...

Interactive Complex Analysis, ver. 1.0.0
(c) 2006-2007, Bert G. Wachsmuth