pre calculus

cube roots of 64i?

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asked by joe
  1. 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-164/64=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.

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  2. β=90° (sin-164/64=sin-11=π/2)

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