Question

Find the locus of a complex number, $z=x+iy$, satisfying the relation $\left|\frac{z-3i}{z+3i}\right|\le \sqrt{2}$
Illustrate the locus of $z$ in the Argand plane.

Answer 1

Let $z=x+iy$
then $\left|\frac{z-3i}{z+3i}\right|\le \sqrt{2}$
$\left|\frac{x+iy-3i}{x+iy+3i}\right|\le \sqrt{2}$
$\left|\frac{x+i(y-3)}{x+i(y+3)}\right|\le \sqrt{2}$
$\sqrt{x^2+{(y-3)}^2}\le \sqrt{2}\sqrt{x^2+{(y+3)}^2}$
On squaring on both sides, we get
$x^2+{(y-3)}^2\le 2(x^2+{(y+3)}^2)$
$x^2+y^2+9-6y\le 2x^2+2y^2+18+12y$
$x^2+y^2+18y+9\ge 0$
A circle with centre $(0,-9)$ and radius $6\sqrt{2}$units.