Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
1  
It's easy to prove that all open half planes (cut through lines through zero, that is) are closed under addition, so you don't need the projection argument. –  Thomas Andrews Aug 22 '12 at 18:00
    
@ThomasAndrews True. In fact I just realized that $P$ is precisely the functional $ax+by$. –  Alex Becker Aug 22 '12 at 18:02
    
Hah, of course! –  Thomas Andrews Aug 22 '12 at 18:03
    
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
    
@ThomasAndrews Thanks, fixed it. –  Alex Becker Aug 22 '12 at 18:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.