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  1. Euphemia
    0
    2021-07-28T18:35:02+00:00
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    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)

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