The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$
-2018 AMC 12A Problem 22

The solution I found online was based on that the triangle formed by the origin and the two complex numbers in the first quadrant is $\frac{1}{4}$ area of the parallelogram, and that can be found since the $z^{2}=2+2\sqrt{3}i$ is easily convertible into polar coordinates and square-rooted with DeMoivre's, but it didn't find the coordinates of the $z^{2}=4+4\sqrt{15}i$.

In the solutions section of the AoPS problem page, however, one of the solutions is this:

The roots are $\pm\left(\sqrt{10}+i\sqrt{6}\right),\pm\left(\sqrt{3}+i\right)$ (easily derivable by using DeMoivre and half-angle). From there, shoelace on $\left(0,0\right),\left(\sqrt{10},\sqrt{6}\right),\left(\sqrt{3},1\right)$ and multiplying by $4$ gives the area of $6\sqrt{2}-2\sqrt{10}$, so the answer is $\boxed{20}$. (trumpeter)

This was what I tried to do at first, before realizing that I have no idea how $\sqrt{15}$ works into the trigonometric representation formulas. Apparently, though, it is "easily derivable by using DeMoivre and half-angle." Using this description of the solution process, how would one go about finding each of the roots of the two complex numbers given in the problem?

  • $\begingroup$ The AoPS YouTube channel posts videos solving the final five problems from each year's AMC 10/12 A and B competitions. Here is the video for this problem: youtube.com/…. Richard Rusczyk is terrific at explaining solutions :) $\endgroup$ – Soham Konar Dec 30 '19 at 2:05
  • $\begingroup$ Yes, Richard is amazing! but I linked to that video too. I'm really just trying to understand the solution by trumpeter. $\endgroup$ – David Dong Dec 30 '19 at 2:09
  • $\begingroup$ I understand your confusion but I haven't looked at the problem or solution too closely yet because I haven't taken the 2018 AMC 12A as practice yet. $\endgroup$ – Soham Konar Dec 30 '19 at 2:20

Let $2\theta$ be an angle that satisfies $$\cos 2\theta = \frac{4}{\sqrt{4^2 + (4\sqrt{15})^2}} = \frac{1}{4}.$$ Then by the double-angle identity $$\cos 2\theta = 2 \cos^2 \theta - 1,$$ we readily obtain $$\cos \theta = \pm \sqrt{\frac{5}{8}}$$ hence $$\sin \theta = \pm \sqrt{\frac{3}{8}},$$ where the signs are chosen to be the same in each case because $z^2$ (hence $2\theta$) is in quadrant I, hence $z$ (and $\theta$) is in quadrant I or III. The rest is straightforward.


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