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(Mandelbrot) There is a unique pair of positive real numbers satisfying the equations \begin{equation} x^4 - 6x^2y^2 + y^4 = 8 \hspace{1em} \text{and} \hspace{1em} x^3y - xy^3 = 2\sqrt{3}\text{.} \end{equation} Determine $x$, writing your answer in the form $a \cos\theta$, with $\theta$ in degrees.

My answer is written below, but I would love to see any easier solutions!

I would like to note that this problem came from a problem set given to me in one of my classes, the day we were learning about complex numbers and De Moivre's theorem. It would be great if any of y'all could find a solution to this problem using complex numbers :D

crxyz
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2 Answers2

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$$(x+iy)^4 = (x^4 - 6x^2y^2 + y^4) + 4i(x^3y - xy^3) $$

we get $(x+iy)^4 = 8 + 8i \sqrt 3.$ As $\frac{8 \sqrt 3}{8} = \sqrt 3$ we see that the angle with the $x$ axis is $\frac{\pi}{3}.$ the angle of our $x +iy$ must then be $\frac{\pi}{12}.$ Lots of questions today about this angle, including sine, cosine, and tangent

Will Jagy
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  • I would also note that the angle of $ x + iy $, after using De Moivre's theorem, could also equal $ \frac{\pi}{12} + \frac{1}{2} k \pi $ where $ k \in \mathbb{Z} $ such that $ 0 \le k < 4 $. But, all possible angles other than $ \frac{\pi}{12} $ result in a negative $x$, so $ \frac{\pi}{12} $ is our answer (you can also check $y$ and find that it's positive). – crxyz Sep 26 '22 at 18:52
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\begin{align*} &x^4 - 6x^2y^2 + y^4 = 8 \\ \implies &(x^2-y^2)^2 - 4x^2y^2 = 8\\ \end{align*} and \begin{align*} &x^3y-xy^3 = 2\sqrt{3} \\ \implies &xy(x^2-y^2) = 2\sqrt{3} \\ \end{align*} Let $x^2 - y^2 = a$, $xy = b$. Then, \begin{align*} a^2-4b^2 &= 8 \\ ab &= 2\sqrt{3} \\ \end{align*} Solving, \begin{align*} &a^2 - 4(2\sqrt{3}/a)^2 = 8 \\ \implies &a^4 - 8a^2 - 48 = 0 \\ \implies &a^2 = 12, -4 \text{, so } a = \pm2\sqrt{3}, \pm 2i \end{align*} Clearly, $x$ and $y$ won't be real if $a = \pm2i$, so $a = \pm2\sqrt{3}$ and $b = \pm 1$. And, because $xy = b$ is positive, $b = 1$, which means $ 2\sqrt{3} / b = a$ is also positive, so $a = 2\sqrt{3}$.
Then, $xy = 1 \implies y = 1/x$ so $x^2 - 1/x^2 = a \implies x^4 - 2\sqrt{3}x^2 - 1 = 0$ Using the quadratic equation, \begin{equation} x = \sqrt{\frac{2\sqrt{3} \pm \sqrt{16}}{2}} = \sqrt{ \sqrt{3} \pm 2} \end{equation} Because $ \sqrt{3} < 2 $, $x = \sqrt{2 + \sqrt{3}}$. Using basic algebra and setting $ x = \sqrt{m} + \sqrt{n} $, we find $ (m, n) = (3/2, 1/2) $. So, $ x = \sqrt{3/2} + \sqrt{1/2} = \frac{\sqrt{6} + \sqrt{2}}{2} $. One can also notice that $ \cos 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4} $, so $ x = 2\cos 15^{\circ}. \hspace{0.5em} \blacksquare$

crxyz
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