非零复数a,b满足a^2+ab+b^2=0,则(a/(a+b))^1999+(b/(a+b))^1999的值是?
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![非零复数a,b满足a^2+ab+b^2=0,则(a/(a+b))^1999+(b/(a+b))^1999的值是?](/uploads/image/z/15040612-28-2.jpg?t=%E9%9D%9E%E9%9B%B6%E5%A4%8D%E6%95%B0a%2Cb%E6%BB%A1%E8%B6%B3a%5E2%2Bab%2Bb%5E2%3D0%2C%E5%88%99%EF%BC%88a%2F%28a%2Bb%29%29%5E1999%2B%EF%BC%88b%2F%28a%2Bb%29%29%5E1999%E7%9A%84%E5%80%BC%E6%98%AF%3F)
非零复数a,b满足a^2+ab+b^2=0,则(a/(a+b))^1999+(b/(a+b))^1999的值是?
非零复数a,b满足a^2+ab+b^2=0,则(a/(a+b))^1999+(b/(a+b))^1999的值是?
非零复数a,b满足a^2+ab+b^2=0,则(a/(a+b))^1999+(b/(a+b))^1999的值是?
复数a,b非零,且
a^2 + ab + b^2 = 0,
所以,
(a+b)^2 = ab
a/(a+b) = (a+b)/b = [b/(a+b)]^(-1) ...(1)
又,
a/(a+b) + b/(a+b) = 1 ...(2)
令 u = a/(a+b),则由(1)和(2)解得,
a/(a+b) = u = (1 + i(3)^(1/2))/2
= exp[iPI/3],
b/(a+b) = (1 - i(3)^(1/2))/2
= exp[-iPI/3]
或者,
a/(a+b) = u = (1 - i(3)^(1/2))/2
= exp[-iPI/3],
b/(a+b) = (1 + i(3)^(1/2))/2
= exp[iPI/3].
所以,总有,
(a/(a+b))^1999 +(b/(a+b))^1999
= {exp[iPI/3]}^1999 + {exp[-iPI/3]}^1999
= exp[iPI(1999/3)] + exp[-iPI(1999/3)]
= exp[iPI/3] + exp[-iPI/3]
= (1 + i(3)^(1/2))/2 + (1 - i(3)^(1/2))/2
= 1