What are the solutions to the equation z^3+1-i=0 ?

The solutions to the equation z^3+1-i=0 are

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z^3+1-i=0

z^{3} + 1 - i = 0

z = \frac{2 ^{\frac{2}{3}}}{2} + i \frac{2 ^{\frac{2}{3}}}{2}, z = \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}, z = \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} - i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}

Solution steps

z^{3} + 1 - i = 0

Move

i

to the right side

z^{3} + 1 = i

Move

1

to the right side

z^{3} = i - 1

For

z^{n} = a

the solutions are

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

For

n = 3, a = i - 1

∣ a ∣ = 2

ar g (a) = - \frac{π}{4} + π

Simplify

z = \frac{2 ^{\frac{2}{3}}}{2} + i \frac{2 ^{\frac{2}{3}}}{2}, z = \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}, z = \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} - i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}

2 x^{5} + 5 x = 2215

12 x^{3} - 26 x^{2} - 12 x = 0

25 x^{4} + 9 x^{2} - 16 = 0

x^{6} = 23

x^{6} = 2 x

What are the solutions to the equation z^3+1-i=0 ?
The solutions to the equation z^3+1-i=0 are
Find the zeros of z^3+1-i=0
The zeros of z^3+1-i=0 are