# A finite sum of reciprocals of complex numbers cannot be confined to a half-plane

Let $z_1,\dots,z_n$ be non-zero complex numbers such that $\sum_{k=1}^n \frac{1}{z_k} = 0$. Prove that for any line $ax+by=0$ passing through the origin, $z_1,\dots,z_n$ cannot all lie in either of the half-spaces $\{ ax+by<0\}$, $\{ ax+by >0\}$. Any help would be greatly appreciated, thanks in advance!

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Note that $z_1,\ldots,z_n$ all lie in the same half-plane iff $\arg(z_1),\ldots,\arg(z_n)$ differ by less than $\pi$, and that $\arg(z)=-\arg(1/z)$. Thus we can substitute $1/z_i$ for each $z_i$, and need only show that if $\sum_i z_i=0$ then not all $z_i$ live in the same half-plane. Suppose all $z_i$ live in the half-plane defined by $ax+by>0$, and let $z_i=x_i+iy_i$. Then $0<\sum_i ax_i+by_i=a\sum_i x_i+b\sum_i y_i=0$, a contradiction.
@ThomasAndrews True. In fact I just realized that $P$ is precisely the functional $ax+by$. –  Alex Becker Aug 22 '12 at 18:02
That edit doesn't quite work, since $\sum z_i$ is not $\sum ax_i+by_i$ but $(\sum x_i) + i(\sum y_i)$ with $a\sum x_i + b\sum y_i > 0$ being the contradiction. But it is just easier to write that the half-plane is closed under addition and prove it for $z_1+z_2$... –  Thomas Andrews Aug 22 '12 at 18:12