Problem:
Let z1β=18+83i,z2β=18+39i, and z3β=78+99i, where i=β1β. Let z be the unique complex number with the properties that z2ββz1βz3ββz1βββ zβz3βzβz2ββ is a real number and the imaginary part of z is the greatest possible. Find the real part of z.
Solution:
Let z2ββz1βz3ββz1ββ=r1βcis(ΞΈ1β), where 0β<ΞΈ1β<180β.
If z is on or below the line through z2β and z3β, then zβz3βzβz2ββ=r2βcis(ΞΈ2β), where 0β<ΞΈ2β<180β. Because r1βcis(ΞΈ1β)β r2βcis(ΞΈ2β)=r1ββ r2ββ cis(ΞΈ1β+ΞΈ2β) is real, it follows that ΞΈ1β+ΞΈ2β=180β, meaning that z1β,z2β,z3β, and z lie on a circle. On the other hand, if z is above the line through z2β and z3β, then zβz3βzβz2ββ=r2βcis(βΞΈ2β), where 0β<ΞΈ2β<180β. Because r1βcis(ΞΈ1β)β r2βcis(ΞΈ2β)=r1ββ r2ββ cis(ΞΈ1ββΞΈ2β) is real, it follows that ΞΈ1β=ΞΈ2β, meaning that z1β,z2β,z3β, and z lie on a circle. In either case, z must lie on the circumcircle of β³z1βz2βz3β, whose center is the intersection of the perpendicular bisectors of z1βz2ββ and z1βz3ββ, namely, the lines y=239+83β=61 and 16(yβ91)=β60(xβ48). Thus the center of the circle is c=56+61i. The imaginary part of z is maximal when z is at the top of the circle, and the real part of z is 56β.
OR
Let z=a+bi, where a and b are real numbers. Then the given expression is
which simplifies to (aβ56)2+(bβ61)2=1928. Thus a+bi lies on the circle centered at 56+61i with radius 1928β. When b is maximal, z is at the top of the circle, and a=56β.
