ComplexNumberBasic AIME Intermediate
2018


Problem - 4120
Suppose that $x$, $y$, and $z$ are complex numbers such that $xy = -80 - 320i$, $yz = 60$, and $zx = -96 + 24i$, where $i$ $=$ $\sqrt{-1}$. Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$. Find $a^2 + b^2$.

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Answer     74

Solution 1


First we evaluate the magnitudes. $|xy|=80\sqrt{17}$, $|yz|=60$, and $|zx|=24\sqrt{17}$. Therefore, $|x^2y^2z^2|=17\cdot80\cdot60\cdot24$, or $|xyz|=240\sqrt{34}$. Divide to find that $|z|=3\sqrt{2}$, $|x|=40\sqrt{34}$, and $|y|=10\sqrt{2}$.This allows us to see that the argument of $y$ is $\frac{\pi}{4}$, and the argument of $z$ is $-\frac{\pi}{4}$. We need to convert the polar form to a standard form. Simple trig identities show $y=10+10i$ and $z=3-3i$. More division is needed to find what $x$ is.\[x=-20-12i\]\[x+y+z=-7-5i\]\[(-7)^2+(-5)^2=\boxed{074}\]


Solution 2
Dividing the first equation by the second equation given, we find that $\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)$. Substituting this into the third equation, we get $z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i$. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of $y$ is the negative of that of $z$, and their magnitudes multiply to $60$. Thus we have $z=\sqrt{-18i}=3-3i$ and $3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i$. To find $x$, we can use the previous substitution we made to find that $x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i$ Therefore, $x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{074}$ 

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