pre calculus
posted by joe
cube roots of 64i?

MathMate
Convert the given number into the form
a+bi = α(cos(β)+i sin(β))
where
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)
β=90° (sin^{1}64/64=1)
The equivalent angles are:
90°
450°
810°
(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
z1=α'(cos(β1)+isin(β1)
=4(cos(30°)+isin(30°))
z2=4(cos(150°)+isin(150°))
z3=4(cos(270°)+isin(270°))
Check by expanding z1^3, z2^3 and z3^3 to get back 64i. 
MathMate
β=90° (sin164/64=sin^{1}1=π/2)
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