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[tex]z^{7}=128i[/tex]

z = ____ + ____ i

Sagot :

LammettHash

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

[tex]z^7 = 128i = 2^7i = 2^7e^{i\frac\pi2}[/tex]

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

(where n = 0, 1, 2, …, 6)

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

[tex]\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)[/tex]

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