What are the three cube roots of -8i? i = √-1

Answer 1

By Demoivre's Theorem

-2(cos 270° / 3 + i sin 270º / 3)

-2(cos 630º / 3 + i sin 630º / 3)

-2(cos 990º / 3 + i sin 990º / 3)

Answer 2

What are the 3 cube roots of -8i ?

Let z³ = -8i. Solve for z.

-8i = 8(-i) = 2³(-i), -i = i³

One root is z₁ = 0 + 2i.

De Moivre’s Theorem is needed to find the other two roots:

z = r(cos(A)+i sin(A))

z^N = [r(cos(A)+i sin(A))]^N = r^N(cos(N*A)+i sin(N*A))

For z³ = r^N(cos(N*A)+i sin(N*A)) = 8 ( 0 + i (-1) ), substitute N=3

to get r³=8, r=+2, cos(3A)=0, and sin(3A)=-1.

Solve for 3A.

Notice that 3A = the angle between the negative imaginary y-axis and the positive real x-axis in the xy cartesian plane.

There are 3 possible values for 3A, which means 3 possible values for A

and 3 possible values for z: z = 2(cos(A)+i sin(A)).

One value for 3A = 270 degrees = 3𝜋/2 radians, with A₁ = 90 degrees = 𝜋/2 radians = the positive imaginary y-axis. This corresponds to the previous solution: z₁ = 2i.

The other two roots, z₂ and z₃ can be found by adding 360 degrees = 2𝜋 radians and 720 degrees = 4𝜋 radians to the 3A angle for z³. All three of these angles correspond to the negative imaginary y-axis and can be used for 3A:

Use 3A₂ = 3𝜋/2 + 2𝜋 radians, for z₂

Use 3A₃ = 3𝜋/2 + 4𝜋 radians, for z₃

To calculate z₂, A₂ = 7𝜋/6 radians, cos(A₂) = -(√3)/2, sin(A)=-1/2

To calculate z₃, A₃ = 11𝜋/6 radians, cos(A₃) = +(√3)/2, sin(A)=-1/2

Z₂ = r(cos A₂ + i sin A₂) = 2( -(√3)/2-i/2) =  -(√3) - i

Z₃ = r(cos A₃ + i sin A₃) = 2(+(√3)/2-i/2) = +(√3) - i 

Check:

Z₂³ = (-(√3)-i)3 = (3 + 2i (√3) -1)(-(√3)-i) = (2 + 2i (√3))(-(√3)-i) 

      = -2*(√3)-2i-6i+2*(√3) = -8i

Z₃³ = (+(√3)-i)3 = (3 - 2i (√3) -1)(+(√3)-i) = (2 - 2i (√3))(+(√3)-i) 

      = 2*(√3)-2i-6i-2*(√3) = -8i

Answer 3

it would be 2i:

2i*2i*2i = 8i³ = 8i²i = 8(-1)i = -8i