If z_1+z_2+·s+z_n=0 and not all z_k lie on a common line…

If $z_1+z_2+\cdots+z_n=0$ and not all $z_k$ lie on a common line through the origin, show that every line through the origin has points of $\{z_k\}$ strictly on both sides.

Step 1 · Set up an arbitrary line through the origin

Let $L$ be the line through $0$ at angle $\alpha$. Rotating by $e^{-i\alpha}$, define $u_k=e^{-i\alpha}z_k$; then $L$ corresponds to the real axis, and "side" is determined by $\text{sgn}(\text{Im}(u_k))$.

Step 2 · Suppose (for contradiction) all points lie on one closed side

Assume $\text{Im}(u_k)\ge0$ for every $k$ (i.e. no point strictly on the negative side). Since $\sum z_k=0$, multiplying by $e^{-i\alpha}$: $\ \sum u_k=0$, so in particular:

$$ \sum_{k=1}^n\text{Im}(u_k)=0 $$

Step 3 · A sum of nonnegative terms equal to zero forces every term to vanish

Since each $\text{Im}(u_k)\ge0$ and they sum to $0$, every $\text{Im}(u_k)=0$ — meaning every $z_k$ lies exactly on the line $L$.

Step 4 · Contradiction with the hypothesis

This contradicts the assumption that not all $z_k$ lie on a common line through the origin (since $L$ was an arbitrary line — if the points avoided the negative side of $L$, Step 3 would force them all onto $L$ itself, i.e. onto one particular line, contradicting the hypothesis).

Step 5 · Conclusion

Since assuming "no point strictly on the negative side" leads to a contradiction, there must be at least one $z_k$ with $\text{Im}(u_k)<0$. By the identical argument applied to $-u_k$ (or directly, since $\sum\text{Im}(u_k)=0$ and not all terms are zero, both a positive and a negative term must occur), there is also at least one $z_j$ with $\text{Im}(u_j)>0$. So $L$ has points of $\{z_k\}$ strictly on both sides.

$$ \boxed{\text{Every line through the origin separates } \{z_1,\ldots,z_n\} \text{ into both open half-planes}} $$

Verification

$z_1=2,\ z_2=-1+i,\ z_3=-1-i$ (sum $=0$, not collinear). Line $L=$ imaginary axis ($\alpha=\frac\pi2$, so $u_k=-iz_k$): $u_1=-2i$ (Im $<0$), $u_2=1+i$ (Im $>0$), $u_3=-1+i$ (Im $>0$) — points on both sides ✓. Line $L=$ real axis ($\alpha=0$): $\text{Im}(z_1)=0,\ \text{Im}(z_2)=1>0,\ \text{Im}(z_3)=-1<0$ — again both sides ✓ (with $z_1$ exactly on the line, which is consistent, not a counterexample).