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  • 如下,復(fù)數(shù)計(jì)算

    如下,復(fù)數(shù)計(jì)算
    復(fù)數(shù)計(jì)算(1+√3i)^100+(1-√3i)^100=?
    (1加根號(hào)3乘以i)的100次方加上(1減去根號(hào)3乘以i)的一百次方=?,i是在根號(hào)3外面的
    數(shù)學(xué)人氣:810 ℃時(shí)間:2020-05-07 21:31:08
    優(yōu)質(zhì)解答
    (1+√3i)^2 = 1-3 + 2√3i = -2 (1-√3i)
    (1-√3i)^2 = 1-3 - 2√3i = -2 (1+√3i)
    (1+√3i)^3 = -2(1-√3i)*(1+√3i) = -2 * ( 1+4)=-2^3
    (1-√3i)^3 = -2(1+√3i)*(1-√3i) = -2 * ( 1+4)=-2^3
    因此
    (1+√3i)^100+(1-√3i)^100
    = (1+√3i)^99*(1+√3i) + (1-√3i)^99*(1-√3i)
    = -2^99 * (1+√3i) - 2^99*(1-√3i)
    = -2^99 * ( 1+1+√3i-√3i )
    = -2^100
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