Z^(7)=128i z = ____ + ____ i

Question

`z^(7)=128i`

z = ____ + ____ i

Answer 1

If z ⁷ = 128i, then there are 7 complex numbers z that satisfy this equation.

`z^7 = 128i = 2^7i = 2^7e^(i\frac\pi2)`

`\implies z=\sqrt[7]{2^7} e^(i\frac17\left(\frac\pi2+2n\pi\right))`

(where n = 0, 1, 2, …, 6)

`\implies z = 2 e^{i\left(\frac\pi{14}+\frac{2n\pi}7\right)}`

`\displaystyle\implies z = 2 \left(\cos\left(\frac\pi{14}+\frac{2n\pi}7\right)+i\sin\left(\frac\pi{14}+\frac{2n\pi}7\right)\right)`