Problem:
The solutions to the equations z2=4+415i and z2=2+23i, where i=−1, form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form pq−rs, 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?
Answer Choices:
A. 20
B. 21
C. 22
D. 23
E. 24 Solution:
Let z=a+bi be a solution of the first equation, where a and b are real numbers. Then (a+bi)2=4+415i. Expanding the left-hand side and equating real and imaginary parts yields
a2−b2=4 and 2ab=415
From the second equation, b=a215, and substituting this into the first equation and simplifying gives (a2)2−4a2−60=0, which factors as (a2−10)(a2+6)=0. Because a is real, it follows that a=±10, from which it then follows that b=±6. Thus two vertices of the parallelogram are 10+6i and −10−6i. A similar calculation with the other given equation shows that the other two vertices of the parallelogram are 3+i and −3−i. The area of this parallogram can be computed using the shoelace formula, which gives the area of a polygon in terms of the coordinates of its vertices (x1,y1),(x2,y2), …,(xn,yn) in clockwise or counter-clockwise order:
In this case x1=10,y1=6,x2=3,y2=1,x3=−10, y3=−6,x4=−3, and y4=−1. The area is 62−210, and the requested sum of the four positive integers in this expression is 20 .
OR
The solutions of z2=4+415i=16cis2θ1 are z1=4cisθ1 and its opposite, with 0<θ1<4π and tan2θ1=15. Then cos2θ1=41, and by the half-angle identities, cosθ1=410 and sinθ1=46. Similarly, the solutions of z2=2+23i=4cosθ2 are z2=2cosθ2 and its opposite, with 0<θ2<4π and tan2θ2=3. Then cosθ2=23 and \sin \theta_{2}=\dfrac{1}
The area of the parallelogram in the complex plane with vertices z1, z2, and their opposites is 4 times the area of the triangle with vertices 0,z1, and z2, and because the area of a triangle is one-half the product of the lengths of two of its sides and the sine of their included angle, it follows that the area of the parallelogram is