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werwe
@werwe
July 2022
1
10
Report
Возвести в степень по формуле Муавра
(-1+
√3i)^9
Извлечь корень
√(2+2
√3i)
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Удачник66
Verified answer
Нужно представить число в тригонометрической форме.
(-1+
√3*i
)^9 = (2*(-1/2 +
√3/2*i
))^9 = (2*(cos 2pi/3 + i*sin 2pi/3))^9 =
= 2^9*(cos (2*9pi/3) + i*sin (2*9pi/3)) = 512*(cos 6pi + i*sin 6pi) = 512(1 + 0) = 512
√(2 + 2
√3*i
) =
√(4*(1/2 +
√3/2*i
)) =
√(4*(cos pi/3 + i*sin pi/3))
Получаем 2 значения
1) 2*(cos pi/6 + i*sin pi/6) =
√3 + i
2) 2*(cos (pi+pi/6) + i*sin (pi+pi/6)) = 2*(cos 7pi/6 + i*sin 7pi/6) = -
√3 - i
1 votes
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Answers & Comments
Verified answer
Нужно представить число в тригонометрической форме.(-1+√3*i)^9 = (2*(-1/2 + √3/2*i))^9 = (2*(cos 2pi/3 + i*sin 2pi/3))^9 =
= 2^9*(cos (2*9pi/3) + i*sin (2*9pi/3)) = 512*(cos 6pi + i*sin 6pi) = 512(1 + 0) = 512
√(2 + 2√3*i) = √(4*(1/2 + √3/2*i)) = √(4*(cos pi/3 + i*sin pi/3))
Получаем 2 значения
1) 2*(cos pi/6 + i*sin pi/6) = √3 + i
2) 2*(cos (pi+pi/6) + i*sin (pi+pi/6)) = 2*(cos 7pi/6 + i*sin 7pi/6) = -√3 - i