复数1+2i的平方根 我设1+2i的平方根=a+bi 答案算出来根号里还有根号 ……

2个回答

  • √ (1+2i)=a+bi.[实数a=?,b=?]

    [√(1+2i)^2=(a+bi)^2.

    1+2i=a^2+2abi+(bi)^2.

    =a^2+2abi-b^2.

    1+2i=a^2-b^2-2abi.

    a^2-b^2=1 (2).

    2=-2ab.

    ab=-1 (2).

    b=-1/a.

    a^2-(-1/a)^2=1.

    a^4-a^2-1=0

    ( a^2-1/2)^2-1/4-1=0.

    (a^2-1/2)^2=5/4,

    a^2-1/2=±√5/2.

    a^2=1/2± √5/2.

    =(1±√5)/2.

    a1^2=(1±√5)/2;

    a1=[√2(1±√5)]/2

    a2=-[√2(1±√5)]/2

    b1=-1/[√2(1±√5)]/2.

    =-2[√2(1±√5)]/(1-5)

    =(1/2)√2(1±√5).

    b2=2[√2(1±√5)/(1-5).

    =-(1/2)(√2(1±√5).

    b1=-1/a=(-+)2√[1±√5)/2]/[(1±√5)]

    因加减符号好打,但减加符号不好打,方法是对的,具体数值请自己认真核对.