I'm asked to prove that if $z_1$, $\ldots$, $z_k$ lie on one side of a straight line through $0$, then $z_1+\cdots+z_k \neq 0$.
In the proof, we let $\theta$ be the angle between the line and the real axis, and let $w = \cos\theta + i\sin\theta$. Then $z_1w^{-1}$, $\ldots$, $z_kw^{-1}$ all lie on one side of the real axis, so the same is true for $z_1w^{-1}+\cdots+z_kw^{-1} = (z_1+\cdots+z_k)w^{-1}$, which shows that $z_1+\cdots+z_k$ lies on the corresponding side of the original line.
But why is it the case that if we divide the various $z_i$s of the original sequence by $w$ then they all get sorted on one side of the real axis? I get that the points all rotate clockwise by $\theta$, but why?
Thank you all in advance.
Simon.
Edit: I think I've answered it myself; see below.
Best Answer
Hint:
It is not difficult to prove the case $k=2$, than use induction.
Or, following your way:
if all $z_i$ lie on the same side of the line $z=\rho e^{i\theta}$ we have
$$z_i=\rho_ie^{i(\theta +\alpha_i)}$$ with $0<\alpha_i<\pi$ or $-\pi<\alpha_i<0$.
So we have:
$z_i e^{-i\theta}=\rho_ie^{i(\theta +\alpha_i)}e^{-i\theta}=\rho_ie^{i \alpha_i}$
which, due to the limits for $\alpha_i$, are all on the same side of the real axis.