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