Problem:
In the complex plane, let A be the set of solutions to z3β8=0 and let B be the set of solutions to z3β8z2β8z+64=0. What is the greatest distance between a point of A and a point of B ?
Answer Choices:
A. 23β
B. 6
C. 9
D. 221β
E. 9+3β Solution:
We solve each equation separately:
We solve z3β8=0 by De Moivre's Theorem. Let z=r(cosΞΈ+isinΞΈ)=rcisΞΈ, where r is the magnitude of z such that rβ₯0, and ΞΈ is the argument of z such that 0β€ΞΈ<2Ο.
We have
z3=r3cos(3ΞΈ)=8(1)
from which
r3=8, so $r= $ 2 . cos(3ΞΈ)=1 sin(3ΞΈ)=0,, so 3ΞΈ=0,2Ο, 4Ο, or ΞΈ=0, 32Οβ, 34Οβ. The set of solutions to z3β8=0 is A={2,β1+3βi,β1β3βi}. In the complex plane, the solutions form the vertices of an equilateral triangle whose circumcircle has center 0 and radius 2 . We solve z3β8z2β8z+64=0 by factoring by grouping. We have
z2(zβ8)β8(zβ8)=0 (z2β8)(zβ8)=0
The set of solutions to z3β8z2β8z+64=0 is B={22β,β22β,8}.
In the graph below, the points in set A are shown in red, and the points in set B are shown in blue. The greatest distance between a point of A and a point of B is the distance between β1Β±3βi to 8 , as shown in the dashed line segments.
By the Distance Formula, the answer is
(β1β8)2+(Β±3ββ0)2β=84β=D)221ββ .
OR
Alternatively, we can solve z3β8=0 by the difference of cubes:
(zβ2)(z2+2z+4)=0
If zβ2=0, then z=2. If z2+2z+4=0, then z=β1Β±3βi by either completing the square or the quadratic formula. The set of solutions to z3β8=0 is A={2,β1+3βi,β1β3βi}.
Following the rest of Solution 1 gives the answer (D)221ββ.
