Автор - halk3431

Помогите плиз добрые люди

Ответ

Проверено экспертом

Автор - oganesbagoyan
z ₁= 1 -i = √2(1/√2 - i*1/√2) =√2(cos(-π/4+2π*k) +isin(-π/4+2π*k)) ;
∛z₁ =∛√2(cos(-π/4+2π*k) +isin(-π/4+2π*k) =(2)^(1/6)*(cos(-π/4+2π*k)/3 +isin(-π/4+2π*k)/3) .
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z
₂= -√3 - i = 2(-√3/2 -i1/2) = 2(cos(-π +π/6) + isin(-π +π/6)) =
2(cos(2π*k-(5/6)*π) + isin(2π*k-(5/6)*π)

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