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

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The discussion revolves around solving the equation involving complex numbers where the conditions include |x+y+z|=21, |x-y|=2√3, and |x|=3√3. It is established that while no distinct real numbers satisfy the equation x²+y²+z²=xy+yz+xz, complex numbers do. The solution involves treating the problem geometrically, identifying that the complex numbers form an equilateral triangle in the complex plane. By applying the cosine rule and calculating distances, the final result is derived as |y|² + |z|² = 132. The discussion highlights the interplay between algebraic manipulation and geometric interpretation in complex number problems.
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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.:)
 
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Peekaboo!

(Sorry, couldn't help myself when I saw that Opalg[/color] was responding. ;))
 
anemone said:
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, $|y|^2 + |z|^2 = 57 + 75 = 132$.
 
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Thank you very much, Opalg for providing a solution so clear.:)
 
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