√ (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.