Question

已知共轭复数Z1与Z2满足(Z1-Z2)2+3Z1Z2i=-8+12i求Z1,Z2

Answer 1

两个实部相等，虚部互为相反数的复数互为共轭复数
设Z1=a+bi 则Z2=a-bi
Z1-Z2=2bi
Z1Z2=a²+b²
代入得：
(2bi)²+3(a²+b²)i=-8+12i
-4b²+3(a²+b²)i=-8+12i
则
b²=2 b=±√2
a²+b²=4
即 a²=2 a=±√2
Z1=±√2(1±i)
对应地
Z2=±√2(1±i)
不懂继续追问

Answer 2

z1=a+bi z2=a-bi
z1-z2=2bi
z1z2=a^2+b^2
所以:(Z1-Z2)^2+3Z1Z2i=(2bi)*2+3(a^2+b^2)i
=-4b^2+3(a^2+b^2)i
=-8 +12i
4b^2=8 b=根号2 or b=-根号2
3(a^2+b^2)=12 a^2+b^2=4 a^2=4-b^2=4-2=2 a=根号2 or a=-根号2
z1=根号2+根号2i z2=根号2-根号2i
or z1=-根号2+根号2i z2=-根号2+根号2i

Answer 3

let
z1=r(cosa+isina)
z2=r(cosa-isina)
(z1-z2)^2+3z1z2i
=r^2(2isina)^2+ 3r^2i
=r^2(-4(sina)^2+3i) =-8+12i
3r^2=12
r=2
-4r^2(sina)^2= -8
(sina)^2=1/2
a=(π/4)
z1=r(cosa+isina)=√2(1+i)
z2=√2(1-i)

Answer 4

设z1，z2分别为x+yi，x-yi，则有(x+yi-x+yi)²+3(x+yi)(x-yi)i=-8+12i，解得x=±√2，y=±√2