Toronto Math Forum
MAT3342020F => MAT334Tests and Quizzes => Quiz 1 => Topic started by: Xinqiao Li on September 25, 2020, 11:17:28 AM

Problem:
Describe the locus of points z satisfying the given equation.
$z+1^2+2z^2=z−1^2.$
Solution:
Let $z=x+yi$ where x,y are real numbers
$z+1^2 = (x+yi)+1^2=(x+1)+yi^2=(x+1)^2+y^2=x^2+2x+1+y^2$
$2z^2=2x+yi^2=2(x^2+y^2)=2x^2+2y^2$
$z1^2 = (x+yi)1^2=(x1)+yi^2=(x1)^2+y^2=x^22x+1+y^2$
Hence,
$z+1^2+2z^2=z−1^2$
$x^2+2x+1+y^2+2x^2+2y^2=x^22x+1+y^2$
$2x^2+4x+2y^2=0$
$2(x^2+2x+y^2)=0$
$x^2+2x+y^2=0$
$x^2+2x+1+y^2=0$
$(x+1)^2+y^2=1$
The locus of point z is a circle centered at (1,0) with radius 1.