Problem: 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 $p+q+r+s?$
$\textbf{(A) } 20 \qquad  \textbf{(B) } 21 \qquad  \textbf{(C) } 22 \qquad  \textbf{(D) } 23 \qquad  \textbf{(E) } 24$

Solution: Let $z_1$ and $z_2$ be the solutions to the equation $z^2=4+4\sqrt{15}i,$ and $z_3$ and $z_4$ be the solutions to the equation $z^2=2+2\sqrt 3i.$ Clearly, $z_1$ and $z_2$ are opposite complex numbers, so are $z_3$ and $z_4.$ This solution refers to the results of De Moivre's Theorem in Solution 2.
From Solution 2, let $z_1=4\operatorname{cis}\phi$ for some $0<\phi<\frac{\pi}{4}.$ It follows that $z_2=4\operatorname{cis}(\phi+\pi).$ On the other hand, we have $z_3=2\operatorname{cis}\frac{\pi}{6}$ and $z_4=2\operatorname{cis}\frac{7\pi}{6}$ without the loss of generality. Since $\tan(2\phi)>\tan\frac{\pi}{3},$ we deduce that $2\phi>\frac{\pi}{3},$ from which $\phi>\frac{\pi}{6}.$
In the complex plane, the positions of $z_1,z_2,z_3,$ and $z_4$ are shown below: [asy] /* Made by MRENTHUSIASM */ size(200);   int xMin = -5; int xMax = 5; int yMin = -5; int yMax = 5; int numRays = 24;  //Draws a polar grid that goes out to a number of circles  //equal to big, with numRays specifying the number of rays:  void polarGrid(int big, int numRays)  {   for (int i = 1; i < big+1; ++i)   {     draw(Circle((0,0),i), gray+linewidth(0.4));   }   for(int i=0;i<numRays;++i)    draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); }  //Draws the horizontal gridlines void horizontalLines() {   for (int i = yMin+1; i < yMax; ++i)   {     draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));   } }  //Draws the vertical gridlines void verticalLines() {   for (int i = xMin+1; i < xMax; ++i)   {     draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));   } }  horizontalLines(); verticalLines(); polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),(2,0)); label("Im",(0,yMax),(0,2));  pair Z1, Z2, Z3, Z4;  Z1 = (sqrt(10),sqrt(6)); Z2 = (-sqrt(10),-sqrt(6)); Z3 = (sqrt(3),1); Z4 = (-sqrt(3),-1);  label("$z_1$", Z1, dir(Z1), UnFill); label("$z_2$", Z2, dir(Z2), UnFill); label("$z_3$", Z3, (0.75,-0.75), UnFill); label("$z_4$", Z4, (-0.75,0.75), UnFill);  draw(Z1--Z3--Z2--Z4--cycle,red);  dot(Z1, linewidth(3.5)); dot(Z2, linewidth(3.5)); dot(Z3, linewidth(3.5)); dot(Z4, linewidth(3.5)); [/asy] Note that the diagonals of every parallelogram partition the shape into four triangles with equal areas. Therefore, to find the area of the parallelogram with vertices $z_1,z_2,z_3,$ and $z_4,$ we find the area of the triangle with vertices $0,z_1,$ and $z_3,$ then multiply by $4.$
Recall that $|z_1|=4, |z_2|=2, \sin\phi=\frac{\sqrt6}{4},$ and $\cos\phi=\frac{\sqrt{10}}{4}$ from Solution 2. The area of the parallelogram is \begin{align*} 4\cdot\left[\frac12\cdot|z_1|\cdot|z_3|\cdot\sin\left(\phi-\frac{\pi}{6}\right)\right] &= 16\sin\left(\phi-\frac{\pi}{6}\right) \\ &= 16\left[\sin\phi\cos\frac{\pi}{6}-\cos\phi\sin\frac{\pi}{6}\right] \\ &= 16\left[\frac{\sqrt3}{2}\sin\phi-\frac12\cos\phi\right] \\ &= 6\sqrt2-2\sqrt{10}, \end{align*} so the answer is $6+2+2+10=\boxed{20}.$

Answer: A 20