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Dимасuk
@Dимасuk
August 2022
2
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Решите уравнение в комплексных числах:
[tex]x^{6} = -1 [/tex]
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IUV
Verified answer
Х^6=-1=e^(i*(pi+2*pi*k))
x=e^(i*(pi+2*pi*k)/6)=cos(pi/6+pi*k/3)+i*sin(pi/6+pi*k/3)
x1=cos(pi/6)+i*sin(pi/6)=корень(3)/2+i*1/2
x2=cos(pi/2)+i*sin(pi/2)=i
x3=cos(5pi/6)+i*sin(5pi/6)=-корень(3)/2+i*1/2
x4=cos(7pi/6)+i*sin(7pi/6)=-корень(3)/2-i*1/2
x5=cos(3pi/2)+i*sin(3pi/2)=-i
x6=cos(11pi/6)+i*sin(11pi/6)=корень(3)/2-i*1/2
3 votes
Thanks 2
Dимасuk
Спасибо)
IUV
на здоровье
iosiffinikov
Verified answer
Ехp(6*i*q)=-1
6*q=pi+2*pi**k
q=pi/6+pi*k/3
x=cos(pi/6+pi*k/3)+i*sin(pi/6+pi*k/3)
Шесть корней :
x=sqrt(3)/2+i*0,5 x=i x=-sqrt(3)/2-i*0,5 x=-i x=-sqrt(3)/2+i*0,5 x=sqrt(3)/2-i*0,5
1 votes
Thanks 1
Dимасuk
Спасибо!)
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Answers & Comments
Verified answer
Х^6=-1=e^(i*(pi+2*pi*k))x=e^(i*(pi+2*pi*k)/6)=cos(pi/6+pi*k/3)+i*sin(pi/6+pi*k/3)
x1=cos(pi/6)+i*sin(pi/6)=корень(3)/2+i*1/2
x2=cos(pi/2)+i*sin(pi/2)=i
x3=cos(5pi/6)+i*sin(5pi/6)=-корень(3)/2+i*1/2
x4=cos(7pi/6)+i*sin(7pi/6)=-корень(3)/2-i*1/2
x5=cos(3pi/2)+i*sin(3pi/2)=-i
x6=cos(11pi/6)+i*sin(11pi/6)=корень(3)/2-i*1/2
Verified answer
Ехp(6*i*q)=-16*q=pi+2*pi**k
q=pi/6+pi*k/3
x=cos(pi/6+pi*k/3)+i*sin(pi/6+pi*k/3)
Шесть корней :
x=sqrt(3)/2+i*0,5 x=i x=-sqrt(3)/2-i*0,5 x=-i x=-sqrt(3)/2+i*0,5 x=sqrt(3)/2-i*0,5