Question

急！已知复数Z=1+i求满足aZ+2bˉZ=（a+2Z)^2的实数a，b的值

Z上面的ˉ

Answer 1

ˉZ=1-i
aZ+2bˉZ=(a+2b)+(a-2b)i
（a+2Z)^2=(a+2+2i)^2=(a+2)^2+4(a+2)i+(2i)^2=(a^2+4a)+4(a+2)i
实部、虚部对应相等a+2b=a^2+4a,a-2b=4(a+2)
两式相加a^2+6a+8=0，a=-2,-4,
a=-2,b=-1; a=-4,b=2

Answer 2

a(1+i)+2b(1-i)=(a+2+2i)^2
a+2b+(a-2b)i=(a+2)^2-4+4(a+2)i
a+2b=(a+2)^2-4
a-2b=4(a+2)
2b= -3a-8
-2a-8=(a+2)^2-4
a^2+6a+8=0
a=-2, -4
b=-(3a+8)/2=-1,2

(a,b)= (-2,-1) or (-4, 2)