Problem:
The solutions to the equations z2=4+415βi and z2=2+23βi, where i=β1β, form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form pqββrsβ, 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+415βi. Expanding the left-hand side and equating real and imaginary parts yields
a2βb2=4 and 2ab=415β
From the second equation, b=a215ββ, and substituting this into the first equation and simplifying gives (a2)2β4a2β60=0, which factors as (a2β10)(a2+6)=0. Because a is real, it follows that a=Β±10β, from which it then follows that b=Β±6β. Thus two vertices of the parallelogram are 10β+6βi and β10ββ6βi. A similar calculation with the other given equation shows that the other two vertices of the parallelogram are 3β+i and β3ββi. 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 (x1β,y1β),(x2β,y2β), β¦,(xnβ,ynβ) in clockwise or counter-clockwise order:
In this case x1β=10β,y1β=6β,x2β=3β,y2β=1,x3β=β10β, y3β=β6β,x4β=β3β, and y4β=β1. The area is 62ββ210β, and the requested sum of the four positive integers in this expression is 20β .
OR
The solutions of z2=4+415βi=16cis2ΞΈ1β are z1β=4cisΞΈ1β and its opposite, with 0<ΞΈ1β<4Οβ and tan2ΞΈ1β=15β. Then cos2ΞΈ1β=41β, and by the half-angle identities, cosΞΈ1β=410ββ and sinΞΈ1β=46ββ. Similarly, the solutions of z2=2+23βi=4cosΞΈ2β are z2β=2cosΞΈ2β and its opposite, with 0<ΞΈ2β<4Οβ and tan2ΞΈ2β=3β. Then cosΞΈ2β=23ββ and \sin \theta_{2}=\dfrac{1}
The area of the parallelogram in the complex plane with vertices z1β, z2β, and their opposites is 4 times the area of the triangle with vertices 0,z1β, and z2β, 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