∗|z+1−2iˉz+3+4i|=1
⇔|z+1−2i|=|ˉz+3+4i|
⇔|(x+1)+(y−2)i|=|(x+3)+(−y+4)i|
⇔(x+1)2+(y−2)2=(x+3)2+(−y+4)2
⇔x−y+5=0 (1)
∗z−2iˉz+i=x+(y−2)ix+(−y+1)i=[x+(y−2)i][x−(−y+1)i]x2+(−y+1)2=x2−y2+3y−2x2+(−y+1)2+2xy−3xx2+(−y+1)2i
Để z−2i¯z+i là số ảo
thì x2−y2+3y−2=0 (2)
* (1), (2) ⇒x=−127,y=237
Vậy z=−127+237i