Understanding a proof that a collection of complex numbers on one side of a line through $0$ must have a non-zero sum 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.
 A: 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.
A: I might have convinced myself; but should I be convinced?
I suspect that it's easier for me to think of this points using polar coordinates:
we started with:
$$x+iy=z=|z|(\cos\theta+i\sin\theta)$$
so
$$x=|z|\cos\theta$$and$$y=|z|\sin\theta$$
in a comment below my initial question I've found that the $\theta$ clockwise rotation of $z$ obtained by $zw^{-1}$ as above is given in terms of $x$ and $y$ by the formula:
$$z''=(x\cos\theta+y\sin\theta)-i(x\sin\theta-y\cos\theta)$$
let $x\gt0$ and we can accept $\theta=arctan (y/x)$, and thus we can substitute as such:
$${x\over \sqrt{y^2+x^2\over x^2}}+{y^2\over x\sqrt{y^2+x^2\over x^2}}-{iyx\over x\sqrt{y^2+x^2\over x^2}}+{iy\over \sqrt{y^2+x^2\over x^2}}={x^2\over\sqrt{y^2+x^2}}+{y^2 \over\sqrt{y^2+x^2}}={x^2\over |z|}+{y^2\over |z|}= \frac 1{|z|}(x^2+y^2)=\frac 1{|z|}{|z|}^2\cos^2\theta+\frac 1{|z|}{|z|}^2\sin^2\theta=|z|(\cos^2\theta+\sin^2\theta)=|z|=|z|(\cos0+i\sin0)$$
which is the point $z''$ with the same modulus $|z|$ of $z$ but laid on the real axis.
makes sense?
A: Got it!
let $$z=|z|(\cos(\theta+|r|)+i\sin(\theta+|r|))$$
so $$zw^{-1}=\frac{|z|(\cos(\theta+|r|)+i\sin(\theta+|r|))}{\cos\theta+i\sin\theta}=|z|\frac{\cos\theta\cos|r|-\sin\theta\sin|r|+i\sin\theta\cos|r|+i\cos\theta\sin|r|}{\cos\theta+i\sin\theta}=|z|\frac{\cos\theta\cos|r|+i^2\sin\theta\sin|r|+i\sin\theta\cos|r|+i\cos\theta\sin|r|}{\cos\theta+i\sin\theta}=|z|\frac{(\cos\theta+i\sin\theta)(\cos|r|+i\sin|r|)}{\cos\theta+i\sin\theta}=|z|(\cos|r|+i\sin|r|)$$
which is the point $z''$ of lenght $|z|$ with an angle of $|r|$ radians.
I feel like this answers the question.
