## Proposition 1.2.18: Roots of Unity |

The |

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

w_{n}= cis(2/n)

Then

w_{n}^{k}= cis(2 k / n)

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

(w_{n}^{k})^{n}= cis(2 k n / n) = cis(2 k ) = 1

for all positive integers *k*. Thus, each *w _{n}^{k}*
is an

As for the second part, let

S_{n}= 1 + z + z^{2}+ z^{3}+ ... + z^{n}

Then

z S_{n}= z + z^{2}+ z^{3}+ ... + z^{n}+ z^{n+1}

Subtracting terms we get

S_{n}- z S_{n}= S_{n}(1-z) = 1 - z^{n+1}

because all terms of *S _{n} - z S_{n}* except the
first and last remain. But then, for

S_{n}= 1 + z + z^{2}+ z^{3}+ ... + z^{n}=^{1-zn+1}/_{1-z}

or equivalently *S _{n-1} = ^{1-zn}/_{1-z}*.
But then

(1-z) S_{n-1}= (1-z)(1 + z + z^{2}+ z^{3}+ ... + z^{n-1}) = 1 - z^{n}

If we now substitute *w _{n} = cis(2/n)*
for

(1-w_{n})(1 + w_{n}+ w_{n}^{2}+ ... + w_{n}^{n-1}) = 1 - (w_{n})^{n}= 0

because *w _{n}^{n} = 1*. But now our statement
follows because

[ x ]

Here are, for example, the eight roots of *z ^{8}=1*:

The 8 roots ofz^{8}=1Sum of 8-th roots is zero

Incidentally, the fact that

(1-z)(1 + z + z^{2}+ z^{3}+ ... + z^{n-1}) = 1 - z^{n}

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