Problem:
If the six solutions of x6=−64 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. −2
B. 0
C. 2i
D. 4
E. 16
Solution:
Use the polar form, x=r(cosθ+isinθ). By DeMoivre's Theorem, r6(cos6θ+isin6θ)=x6=−64=26(cos(180∘+360∘k)+isin(180∘+360∘k)).
Thus r=2 and, using k=−3,−2,−1,0,1,2, we have
θ=(180∘+360∘k)/6=±30∘,±90∘,±150∘.
Since a>0,θ=±30∘, so x=2(cos(±30∘)+isin(±30∘))=3±i. The product of these two roots is (3+i)(3−i)=4.
OR
Recall, from DeMoivre's Theorem, that the six sixth roots of −64 are equispaced around the circle of radius 664. Since ±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 (664)2=4.