What are the solutions to the equation z^3=27i ?

The solutions to the equation z^3=27i are z=(3sqrt(3))/2+i 3/2 ,z=-(3sqrt(3))/2+i 3/2 ,z=-3i

Find the zeros of z^3=27i

The zeros of z^3=27i are z=(3sqrt(3))/2+i 3/2 ,z=-(3sqrt(3))/2+i 3/2 ,z=-3i

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z^{3} = 27 i

z = \frac{3 3}{2} + i \frac{3}{2}, z = - \frac{3 3}{2} + i \frac{3}{2}, z = - 3 i

Solution steps

z^{3} = 27 i

For

z^{n} = a

the solutions are

k = 0, 1, \dots, n - 1

For

n = 3, a = 27 i

∣ a ∣ = 27

ar g (a) = \frac{π}{2}

Simplify

z = \frac{3 3}{2} + i \frac{3}{2}, z = - \frac{3 3}{2} + i \frac{3}{2}, z = - 3 i

(x^{2} + 4) (x + 1) = 0

0 = x^{3} - 4 x^{2} - 9 x + 36

3^{10} = 3^{7} \cdot x^{3}

x^{3} + 4 x^{2} - 4 x - 16 = 0

x^{3} + 2 x^{2} - 16 x - 32 = 0

What are the solutions to the equation z^3=27i ?
The solutions to the equation z^3=27i are z=(3sqrt(3))/2+i 3/2 ,z=-(3sqrt(3))/2+i 3/2 ,z=-3i
Find the zeros of z^3=27i
The zeros of z^3=27i are z=(3sqrt(3))/2+i 3/2 ,z=-(3sqrt(3))/2+i 3/2 ,z=-3i