If $x$ is a complex number and $x^N = 1, x^2\not = 1$, is it true that $\sum_{n=0}^{N-1} x^{2n}=0$?

Question

If $x$ is a complex number and $x^N = 1, x^2\not = 1$, is it true that $\sum_{n=0}^{N-1} x^{2n}=0$?

We know a formula $1+x + x^2 + \dots + x^{N-1} = \frac{1-x^N}{1-x}$.

Answer 1

Change $x$ to $x^2$ in $$1+x + x^2 + \dots + x^{N-1} = \frac{1-x^N}{1-x}$$

to get $$1+x^2 + x^4+ \dots + x^{2N-2} = \frac {1-x^{2N}}{1-x^2}$$

Cross multiply to get

$$ 1-x^{2N} = (1+x^2 + x^4+ \dots + x^{2N-2})(1-x^2)$$

Since $ 1-x^{2N}=0$ and $1-x^2 \ne 0$ we get $1+x^2 + x^4+ \dots + x^{2N-2}=0$

Answer 2

If $x^N = 1$, then $x = e^{\frac{2k\pi i}{N}}$ for some $k$. But then $$\sum_{n=0}^{N-1} x^{2n} = \sum_{n=0}^{N-1} \left(e^{\frac{4k\pi i}{N}} \right)^n = \frac{1-\overbrace{\left(e^{\frac{4k\pi i}{N}} \right)^N}^{=1}}{1-e^{\frac{4k\pi i}{N}}} =0.$$ So, in short: yes.

Edit: $x^2 \ne 1$ is necessary to assure that the demoninator is non-zero.