What are the solutions to the equation z^5=1-i ?

The solutions to the equation z^5=1-i are

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z^5=1-i

z^{5} = 1 - i

z = \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4}, z = \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4} + i \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4}, z = - \frac{2 ^{\frac{3}{5}}}{2} + i \frac{2 ^{\frac{3}{5}}}{2}, z = - \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4}, z = \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4}

Solution steps

z^{5} = 1 - i

For

z^{n} = a

the solutions are

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

For

n = 5, a = 1 - i

∣ a ∣ = 2

ar g (a) = - \frac{π}{4}

Simplify

z = \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4}, z = \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4} + i \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4}, z = - \frac{2 ^{\frac{3}{5}}}{2} + i \frac{2 ^{\frac{3}{5}}}{2}, z = - \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4}, z = \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4}

(1 - p)^{15} = 0

12 x^{3} + 24 x^{2} - 12 x - 24 = 0

4 x^{3} - 48 x^{2} = 0

x^{3} - 3 x^{2} - 10 x - 24 = 0

n^{8} + 12 = 64

What are the solutions to the equation z^5=1-i ?
The solutions to the equation z^5=1-i are
Find the zeros of z^5=1-i
The zeros of z^5=1-i are