Take $$\sum_{k=0}^9 z^k = 0$$ I need to find the smallest positive argument possible for the complex number z. Here's what I figured out so far: So since this is a GP statement, I will have the following where $$z =re^{i\theta}$$:
$$re^{9i\theta} + re^{8i\theta} + re^{7i\theta} + re^{6i\theta} + re^{5i\theta} + re^{4i\theta} + re^{3i\theta} + re^{2i\theta} + re^{i\theta} + 1 = 0$$ Does that mean $$cos9\theta + cos8\theta + cos7\theta + cos6\theta + cos5\theta + cos4\theta + cos3\theta + cos2\theta + cos\theta + 1 = 0$$
PS: The 1 came from when k=0, and hence equals to 1 and belong to the real portion which is the cos
and $$sin9\theta + sin8\theta + sin7\theta + sin6\theta + sin5\theta + sin4\theta + sin3\theta + sin2\theta + sin\theta = 0$$? Is this correct? Am I heading towards a correct approach/direction to this question? Thanks.
Your equation is $$\frac{z^{10}-1}{z-1}=0.$$ Its roots are the tenth roots of unity, apart from $1$, namely $\exp(2n\pi i/10)$ for $n\in\{1,2,\ldots,9\}$.