Just want to know if I'm on the right track here.
let $z=iy$
$|z-4| = 4|z|$
$\Rightarrow \sqrt{(x-4)^2 +y^2} = 4\sqrt{x^2 + y^2}$
$\Rightarrow (x-4)^2 +y^2 = 16x^2 + 16y^2$
$\Rightarrow x^2 -8x -16 -16x^2 = 15y^2$
$\Rightarrow -x^2 -\frac{8}{15}x -\frac{16}{15} = y^2$
$\Rightarrow \pm \sqrt{-x^2 -\frac{8}{15}x -\frac{16}{15} }= y$
Is the locus the set of points on the boundary of this closed shape on the plane? I think I could have used polar co-ordinates here, but is my intuition correct?
Thanks.
$$x^2 +\frac{8}{15}x -\frac{16}{15} + y^2=0$$
this is an equation of a circle:
$$ (x+{4\over 15})^2+y^2 = {16\over 225}+{16\over 15}$$