For any complex $z$, $|z-1|\leq |z-j|+|z-j^2|$

Question

Let $z\in \mathbb C$. Prove that $|z-1|\leq |z-j|+|z-j^2|$

This inequality appears as an exercise in a book for highschoolers. It is marked as very difficult. $j=\exp(2i\pi/3)$ denotes a third root of unity.

I tried squaring both sides, and making use of $j^2=-1-j$, as well as $\bar j=j^2$ but I haven't made significant progress...

Remark: In retrospect it seems obvious that the problem cannot be solved without geometric insights, that I didn't know of at the time.

score 11 · Answer 1 · edited Jul 11 '20 at 09:32

11

Let $A\sim 1$, $B\sim j$ and $C\sim j^2$ then $ABC$ is an equilateral triangle in $\mathbb{C}$. Let $D\sim z$. Then by Ptolemy's inequality we have:

$$AD\cdot BC\leq AB\cdot CD+AC\cdot BD$$

Since $BC=AB = AC$ and $AD =|z-1|$, $CD=|z-j^2|$, $BD =|z-j|$ we get what we want.

edited Jul 11 '20 at 09:32

Martin R

128,226

answered Jul 11 '20 at 08:46

nonuser

91,557

score 11 · Accepted Answer · answered Jul 11 '20 at 09:29

11

(The following is equivalent to Aqua's elegant answer. I'm just stating it to demonstrate a self-contained solution.)

Start with the identity $$ (z-1)(j-j^2) = (z-j)(1-j^2) + (z-j^2)(j-1) \, . $$ Then apply the triangle inequality and use that $$ |j-j^2| = |1-j^2| = |j-1| \, . $$

answered Jul 11 '20 at 09:29

Martin R

128,226

Notice that triangle inequality is equivalent to Ptolemy's inequality (proof goes via inversion) – nonuser Jul 11 '20 at 09:55
@Aqua: Yes, that's why I explicitly stated that this is equivalent to your answer! – Martin R Jul 11 '20 at 09:55

For any complex $z$, $|z-1|\leq |z-j|+|z-j^2|$

2 Answers2

Linked