If some complex numbers lie on the one side of a line which goes through 0, then the sum of them are not 0 Let me describe the problem more specifically.
Suppose that $z_{1},z_{2},\cdots,z_{n}$ are all complex number and they all lie on one side of a straight line passing through $0$. Then $z_{1}+z_{2}+\cdots+z_{n}\neq0$.
I am asked to prove it analytically, even thought it seems quite clear geometrically. So I am thinking that I could use some inequality to prove, but don't have any idea yet..
Thank you very much for any hints or explanations!
 A: Well, you have to give a characterization of those numbers. Since the lines passes through $0$, let's suppose it forms with the positive $x$ semiaxis and angle $\alpha$, what can you say of the $\arg(z_i)$ for all the points on one side and what for those on the other side?
A: Hint:


*

*Choose a (nonzero) factor $c$ such that all numbers $w_i = c \cdot z_i$
lie in the upper half-plane.

*Consider the sum of imaginary parts of the $w_i$.


Detailed solution:
A straight line though the origin with “direction” $\varphi$ is 
given by
$$
 L = \{ te^{i\varphi} \mid t \in \Bbb R \}
 = \{ z \in \Bbb C \mid \operatorname{Im}(e^{-i\varphi} z) = 0 \}
$$
If all $z_i$ lie on one side of the line $L$ then either
$$ 
  \operatorname{Im}(e^{-i\varphi} z_i) > 0 \text{ for all }i
$$
or 
$$ 
  \operatorname{Im}(e^{-i\varphi} z_i) < 0 \text{ for all }i
$$
In both cases,
$$
 \operatorname{Im}(e^{-i\varphi} (z_1 + \ldots + z_n)) \ne 0 \, .
$$
A: Rotate the plane about the origin so the given line coincides with the imaginary axis.  Then all the points have positive real part or all have negative real part.
