Question

复数x=1+i,实数a,b满足a2+2b2=、(a+5)2,求a+么1+})

Answer 1

(1)
let z=a+bi
√3z^2 -11 =(5√3-z^2) i
√3(a^2-b^2)-11 +2√3abi = ((5√3-a^2+b^2-2abi)i
√3(a^2-b^2)-11 =2ab (1)
2√3ab = 5√3-a^2+b^2 (2)
from (1) (2)
√3(a^2-b^2)-11 = (5√3-a^2+b^2 )/√3
3(a^2-b^2)-11√3 = 5√3-a^2+b^2
a^2-b^2 = 4√3
|z|= (a^2-b^2)^(1/2) = 2. 3^(1/4)

(2)
|z1|=|z2|=1
let z1=a1+b1i
z2=a2+b2i
a1^2-b1^2=1 (3)
a2^2-b2^2=1 (4)

z1+z2=i
(a1+a2)+(b1+b2)i=i
a1+a2=0 (1) and
b1+b2=1 (2)
(3)-(4)
a1^2-a2^2 -(b1^2-b2^2) =0
b1^2-b2^2 =0 ( a1 = -a2 from (1) )
b1^2+(1-b1)^2 = 0 ( from (2): b2=1-b1)
1-2b1=0
b1= 1/2
b2=1/2
a1= √5/2 or - √5/2
when a1= √5/2 , a2=-√5/2
when a1= -√5/2 , a2=√5/2

z1z2
=(a1+b1i).(a2+b2i)
= (a1b2-b1b2) + (a1b2+a2b1)i
= (√5-1)/4+ [(√5+1)/4] i or (-√5-1)/4 + [(-√5-1)/4]i