Problem:
If the six solutions of x6=β64 are written in the form a+bi, where a and b are real, then the product of those solutions with a>0 is
Answer Choices:
A. β2
B. 0
C. 2i
D. 4
E. 16
Solution:
Use the polar form, x=r(cosΞΈ+isinΞΈ). By DeMoivre's Theorem, r6(cos6ΞΈ+isin6ΞΈ)=x6=β64=26(cos(180β+360βk)+isin(180β+360βk)).
Thus r=2 and, using k=β3,β2,β1,0,1,2, we have
ΞΈ=(180β+360βk)/6=Β±30β,Β±90β,Β±150β.
Since a>0,ΞΈ=Β±30β, so x=2(cos(Β±30β)+isin(Β±30β))=3βΒ±i. The product of these two roots is (3β+i)(3ββi)=4.
OR
Recall, from DeMoivre's Theorem, that the six sixth roots of β64 are equispaced around the circle of radius 664β. Since Β±2i are roots, exactly two of the roots are in the right half-plane and they must be conjugates. The product of any pair of conjugates is the square of their distance from the origin, so the product of these two roots is (664β)2=4.