Show that if $z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 0 then $z$ $\neq$ 1 and $z^{n+1}$ =1.

Question

Show that if $z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 0 then $z$ $\neq$ 1 and $z^{n+1}$ =1.

I thought I had managed to show that $z$ $\neq$ 1 since if $z$ = 1 then

$z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 1 + 1 ... 1 + 1 = n $\neq$ 0.

But then when I tried to prove the second part of the question I got a contradiction as follows:

Factorising $z^{n+1}$ - 1 = 0 gives ($z$ - 1)($z^n$ + $z^{n-1}$ + ... + $z$ + 1) = 0.

But since $z$ $\neq$ 1 how does this work?

Answer 1

It comes down to the high-school factorisation formula, used to compute the sum of the first $n$ terms of a geometric progression: $$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+\dots+x+1).$$

Answer 2

Write $(z - 1)(z^n + z^{n-1} + ... + z + 1) = 0$, using the null product law and since $z \ne 1$, $(z^n + z^{n-1} + ... + z + 1) = 0$.

Using the formula for $(z^n + z^{n-1} + ... + z + 1)$, we get $$\frac{1-z^{n+1}}{1-z} = 0$$

Hence $z^{n+1} = 1$.