What are the solutions to the equation (2-i)/(1+2i)*z^4=1 ?

The solutions to the equation (2-i)/(1+2i)*z^4=1 are

Find the zeros of (2-i)/(1+2i)*z^4=1

The zeros of (2-i)/(1+2i)*z^4=1 are

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\frac{2 - i}{1 + 2 i} \cdot z^{4} = 1

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

Solution steps

\frac{2 - i}{1 + 2 i} z^{4} = 1

Simplify

\frac{2 - i}{1 + 2 i} : - i

(- i) z^{4} = 1

Divide both sides by

- i

z^{4} = i

For

z^{n} = a

the solutions are

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

For

n = 4, a = i

∣ a ∣ = 1

ar g (a) = \frac{π}{2}

Simplify

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

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

x^{3} - 40 x^{2} + 404 x - 80 = 0

(x + 2) (x + 3) (x + 8) (x + 12) = 54 x^{2}

8 x^{3} + 64 x^{2} + 120 x = 0

c^{3} - 8 = 0

What are the solutions to the equation (2-i)/(1+2i)*z^4=1 ?
The solutions to the equation (2-i)/(1+2i)*z^4=1 are
Find the zeros of (2-i)/(1+2i)*z^4=1
The zeros of (2-i)/(1+2i)*z^4=1 are