MHB Evaluate |y²|+|z²| for Complex Numbers w/ |x+y+z|=21, |x-y|=2√3, |x|=3√3

[anemone]

Hi MHB,

Problem:

While there do not exist pairwise distinct real numbers $x, y, z$ satisfying $x^2+y^2+z^2=xy+yz+xz$, there do exist complex numbers with that property. Let $x, y, z$ be complex numbers such that $x^2+y^2+z^2=xy+yz+xz$ and $|x+y+z|=21$.

Given that $|x-y|=2\sqrt{3}$, $|x|=3\sqrt{3}$, compute $|y^2|+|z^2|$.

Attempt:

I don't believe we have to assign all those variables $a, b, c, d, e, f$ so that we have $x=a+bi$, $y=c+di$ and $z=e+fi$ in order to begin attacking the problem, thus I don't know how to solve it.

Any help would be very much appreciated. Thanks.:)

 

[I like Serena]

Peekaboo!

(Sorry, couldn't help myself when I saw that Opalg[/color] was responding. ;))

 

[Opalg]

While there do not exist pairwise distinct real numbers $x, y, z$ satisfying $x^2+y^2+z^2=xy+yz+xz$, there do exist complex numbers with that property. Let $x, y, z$ be complex numbers such that $x^2+y^2+z^2=xy+yz+xz$ and $|x+y+z|=21$.

Given that $|x-y|=2\sqrt{3}$, $|x|=3\sqrt{3}$, compute $|y^2|+|z^2|$.

Another innocent-looking problem that took a long time to solve! But I see ILS is breathing down my neck so I had better respond fast.

Start in the most naive way possible, by writing $x^2+y^2+z^2=xy+yz+xz$ as a quadratic equation in $x$: $x^2 - (y+z)x + (y^2-yz+z^2) = 0$, with solution $$x = \tfrac12\bigl(y+z \pm\sqrt{-3(y-z)^2}\bigr) = \tfrac12\bigl(y+z \pm3(y-z)i\bigr) = \tfrac{1\pm\sqrt3i}2y + \tfrac{1\mp\sqrt3i}2z = \omega y + \overline{\omega}z,$$ where $\omega = \frac12(1\pm\sqrt 3i) = e^{\pm i\pi/3}$ and the bar denotes complex conjugate. By drawing the sixth roots of unity on the unit circle of the complex plane, you can see that $\omega^2 = \omega - 1 = -\overline{\omega}.$ From those relations, you can check that $y = \omega z + \overline{\omega}x$ and $z = \omega x + \overline{\omega}y.$ Also, $$x-y = (\omega y + \overline{\omega}z) - y = (\omega-1)y + \overline{\omega}z = \overline{\omega}(z - y).$$ It follows that $|x-y| = |z-y|$, and in the same way each of those is equal to $|x-z|$.

Thus $|y-z| = |z-x| = |x-y| = 2\sqrt3.$ So the points $x$, $y$, $z$ form an equilateral triangle in the complex plane, whose centroid is the point $c = \frac13(x+y+z)$ and thus $|c| = 7.$ The triangle has sides of length $2\sqrt3$, and the distance from each vertex to the centroid is $2$.

It's easier to describe what follows if I use capital letters to denote points in the complex plane corresponding to the above complex numbers. So $O$ will denote the origin, and we have an equilateral triangle $XYZ$ with centroid $C$, such that $OX = 3\sqrt3$, $OC = 7$, $XC=2$ and $XY = YZ = ZX = 2\sqrt3.$ (Draw a sketch to see what is going on.)

Now apply the cosine rule in the triangle $OXC$ to find the angle $\theta = \angle OXC$. It tells you that $$\cos\theta = \frac{OX^2 + XC^2 - OC^2}{2OX\cdot XC} = \frac{27 + 4 - 49}{12\sqrt3} = \frac{-18}{12\sqrt3} = -\frac{\sqrt3}2.$$ Therefore $\theta = 5\pi/6$. but $XC$ bisects the $(\pi/3)$-angle of the equilateral triangle. So the angles $OXY$ and $OXZ$ are $\frac56\pi \pm \frac{\pi}6$. In other words one of them, say $OXZ$, is $\pi$, and the other one, $OXY$, is $2\pi/3$.

It follows that $|z| = OZ = OX + XZ = 3\sqrt3 + 2\sqrt3 = 5\sqrt3.$ To find $|y|$, use the cosine rule again, in the triangle $OXY$. That gives $$|y|^2 = OY^2 = OX^2 + XY^2 -2OX\cdot XY \cos(2\pi/3) = 27 + 12 + 2\cdot\tfrac12\cdot 3\sqrt3\cdot 2\sqrt3 = 57.$$ Finally, $|y|^2 + |z|^2 = 57 + 75 = 132$.

 

[anemone]

Thank you very much, Opalg for providing a solution so clear.:)