Question

If |z-1|=2|z+1|then prove that the locus of the point z in the argand plane is 3x2 +3y2 +10x +3=0

Answer 1

$let \: \: z = x + iy \\ |z - 1| = 2 |z + 1| \\ | x + iy- 1| = 2 |x + iy + 1| \\ |(x - 1) + iy| = 2 |(x + 1) + iy| \\ \sqrt{ {(x - 1)}^{2} + {y}^{2} } = 2 \sqrt{ {(x + 1)}^{2} + {y}^{2} } \\ {(x - 1)}^{2} + {y}^{2} = 4({(x + 1)}^{2} + {y}^{2} ) \\ {x}^{2} - 2x + 1 + {y}^{2} = 4 {x}^{2} + 8x + 4 + 4 {y}^{2} \\ 3 {x}^{2} + 3 {y}^{2} + 10x + 3 = 0$