Question

If $|z-2i|+|z-2|\ge ||z|-|z-2-2i||$, then locus of $z$ is

• A. circle
• B. line segment
• C. point
• D. complete complex plane

Answer 1

The correct option is D complete complex plane

Lets write $z=x+iy$ in the given equation:

$|z-2|+|z-2i|\ge ||z|-|z-2-2i||$

$\sqrt{(x-2{)}^2+y^2}+\sqrt{x+(y-2{)}^2}\ge |\sqrt{x^2+y^2}+\sqrt{(x-2{)}^2+(y-2{)}^2}|$

Squaring on both sides, (note that the sign will not be affected during this process)

$(x-2{)}^2+y^2+x^2+(y-2{)}^2+2\sqrt{\left(\right(x-2{)}^2+y^2\left)\right(x^2+(y-2{)}^2)}\ge x^2+y^2+(x-2{)}^2+(y-2{)}^2-2\sqrt{(x^2+y^2)\left(\right(x-2{)}^2+(y-2{)}^2)}$

Cancelling out the common terms on both sides,

$⟹\sqrt{\left(\right(x-2{)}^2+y^2\left)\right(x^2+(y-2{)}^2)}\ge -\sqrt{(x^2+y^2)\left(\right(x-2{)}^2+(y-2{)}^2)}$

Square root of the L.H.S yields us a postive number which is true for R.H.S.

$\Rightarrow (+ve)$ number $\ge -\ast (+ve)$ number

$\Rightarrow (+ve)$ number $\ge (-ve)$ number

which is always true, and

Therefore, $z$ is the complete complex plane.