# If you have four complex numbers and divide each of them by the others then will one of the results have to have non negative real and imaginary parts

Say you have four distinct nonzero complex numbers and you divide each one by the other three (ie you take $$\frac{z_k}{z_j}$$ for each $$k \neq j$$). Is there a way to prove that at least one of these quotients must produce a complex number with nonnegative real and imaginary parts? I've been trying to come up with a counter example all day and I can't find one so I'm wondering if this is something that can actually be proven.

Assume not. If any $$z_i/z_j$$ is in the fourth quadrant, then $$z_j/z_i$$ is in the first. So all quotients must be in the 2nd and 3rd quadrant. Then of the three $$z_1/z_i$$, two must be in the same quadrant, i.e., their quotient $$z_i/z_j$$ must be in the 4th or 1st, contradiction.
The idea is to use a pigeonhole principle type argument: Two of the four numbers must have arguments which differ by at most $$\pi/2$$.
Let $$\alpha_j$$ be the arguments of the four numbers in the range $$[0, 2 \pi)$$ and sort the numbers by their arguments in increasing order. The non-negative differences $$\alpha_2 - \alpha_1, \,\alpha_3-\alpha_2, \,\alpha_4-\alpha_3, \,(2\pi+\alpha_1) -\alpha_4$$ add up to $$2 \pi$$, so that at least one of them must $$\le \pi/2$$. Then the corresponding quotient, that is, one of $$z_2/z_1,\, z_3/z_2, \,z_4/z_3, \,z_1/z_4 \,$$ has an argument between $$0$$ and $$\pi/2$$, which means that it has non-negative real and imaginary part.
• @k12345: You are welcome. – This is a pigeonhole principle type argument: Two of the four numbers must have arguments which differ by at most $\pi/2$. Commented Feb 14, 2021 at 22:21