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

The solutions to the equation z^3=i are z=(sqrt(3))/2+i 1/2 ,z=-(sqrt(3))/2+i 1/2 ,z=-i

Find the zeros of z^3=i

The zeros of z^3=i are z=(sqrt(3))/2+i 1/2 ,z=-(sqrt(3))/2+i 1/2 ,z=-i

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z^{3} = i

z = \frac{3}{2} + i \frac{1}{2}, z = - \frac{3}{2} + i \frac{1}{2}, z = - i

Solution steps

z^{3} = i

For

z^{n} = a

the solutions are

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

For

n = 3, a = i

∣ a ∣ = 1

ar g (a) = \frac{π}{2}

Simplify

z = \frac{3}{2} + i \frac{1}{2}, z = - \frac{3}{2} + i \frac{1}{2}, z = - i

x^{6} + 1 = 0

x^{3} - 7 x + 6 = 0

2 5^{150} = 25 x^{50}

2 x^{3} + x^{2} - 2 x + 1 = 0

x^{3} + 64 = 0

What are the solutions to the equation z^3=i ?
The solutions to the equation z^3=i are z=(sqrt(3))/2+i 1/2 ,z=-(sqrt(3))/2+i 1/2 ,z=-i
Find the zeros of z^3=i
The zeros of z^3=i are z=(sqrt(3))/2+i 1/2 ,z=-(sqrt(3))/2+i 1/2 ,z=-i