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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.

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2 Answers 2

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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.

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    $\begingroup$ This was gorgeous thank you very much $\endgroup$
    – k12345
    Commented Feb 14, 2021 at 22:07
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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.

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  • $\begingroup$ Thank you! This is an interesting way to think about it $\endgroup$
    – k12345
    Commented Feb 14, 2021 at 22:08
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    $\begingroup$ @k12345: You are welcome. – This is a pigeonhole principle type argument: Two of the four numbers must have arguments which differ by at most $\pi/2$. $\endgroup$
    – Martin R
    Commented Feb 14, 2021 at 22:21
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    $\begingroup$ Pigeonhole was definitely not my first thought when I started this problem. Very clever $\endgroup$
    – k12345
    Commented Feb 14, 2021 at 22:22

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