posted by joe
cube roots of 64i?
Convert the given number into the form
a+bi = α(cos(β)+i sin(β))
α = sqrt(a^2+b^2)
β = sin-1(b/α)
If the number is plot in the complex (Z) plane, it will be more evident.
For example, 64i will have
α=64 (the distance from origin)
The equivalent angles are:
(we only need 3)
Divide each of these angles by , to get β1, β2, & β3. And take α'=α^(1/3)=64^(1/3)=4
The three cube roots are then
Check by expanding z1^3, z2^3 and z3^3 to get back 64i.