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Algebra Difficulty 4.2 AIME Find the answer

Suppose that xx, yy, and zz are complex numbers such that xy=80320ixy = -80 - 320i, yz=60yz = 60, and zx=96+24izx = -96 + 24i, where ii == 1\sqrt{-1}. Then there are real numbers aa and bb such that x+y+z=a+bix + y + z = a + bi. Find a2+b2a^2 + b^2.

A number or a short expression. Spacing and $ signs are ignored.

Solution

The First (pun intended) thing to notice is that xyxy and zxzx have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is e3πi2e^{\frac{3\pi i}{2}}. Anyway this results in getting that (3i10)y=z\left(\frac{-3i}{10}\right)y=z. Then substitute this into yzyz to get, after some calculation, that y=10e5πi42y=10e^{\frac{5\pi i}{4}}\sqrt{2} and z=3e7πi42z=-3e^{\frac{7\pi i}{4}}\sqrt{2}. Then plug zz into zxzx, you could do the same thing with xyxy but zxzx looks like it's easier due to it being smaller. Anyway you get x=20+12ix=20+12i. Then add all three up, it turns out easier than it seems because for zz and yy the 2\sqrt{2} disappears after you expand the root of unity (e raised to a specific power). Long story short, you get x=20+12i,y=3+3i,z=1010i    x+y+z=7+5i    a2+b2=074x=20+12i, y=-3+3i, z=-10-10i \implies x+y+z=7+5i \implies a^2+b^2=\boxed{074}.
~First

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.