Question

Locate the complex numbers $z=x+iy$ such that
$|z-1|+|z+1|\le 4$
Show that the point $z$ satisfying the above equation represents the interior and boundary ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$.

Answer 1

$\sqrt{{(x-1)}^2+y^2}+\sqrt{{(x+1)}^2+y^2}\le 4$
Consider $L+M=4$
$L^2-M^2=-4x\therefore L-M=-x$
$\therefore 2L=4-x$ or $4L^2=(4-x{)}^2$
or $4[x^2+y^2-2x+1]=16+x^2-8x$
$3x^2+4y^2=12$.
Hence the given equation is
$\frac{x^2}{4}+\frac{y^2}{3}-1\le 0$ or $S^{\prime }\le 0$ ......$\left(1\right)$
$\frac{x^2}{4}+\frac{y^2}{3}-1$.