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ChuvakZZZ
@ChuvakZZZ
August 2022
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Решить уравнение z^3-i=0
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lolo371
Извлечем корень третьей степени из -i
z=-i = cos(-pi/2) + i*sin(-pi/2)
z^(1/3) = (cos(-pi/2) + i*sin(-pi/2))^(1/3)
z1 = cos(-pi/6) + i*sin(-pi/6) = sqrt(3) / 2 -i *1/2
z2 = cos((-pi/2 +2pi)/3) + i*sin((-pi/2+2pi)/3) = i
z3 = cos((-pi/2 +4pi)/3) + i*sin((-pi/2+4pi)/3) = -sqrt(3)/2 - i*1/2
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Answers & Comments
z=-i = cos(-pi/2) + i*sin(-pi/2)
z^(1/3) = (cos(-pi/2) + i*sin(-pi/2))^(1/3)
z1 = cos(-pi/6) + i*sin(-pi/6) = sqrt(3) / 2 -i *1/2
z2 = cos((-pi/2 +2pi)/3) + i*sin((-pi/2+2pi)/3) = i
z3 = cos((-pi/2 +4pi)/3) + i*sin((-pi/2+4pi)/3) = -sqrt(3)/2 - i*1/2