Question

Let $z$ be a complex number satisfying $|z-3|\le |z-2|,|z-3|\le |z-6|,|z-i|\le |z+i|$ and $|z-i|\le |z-5i|$. Then the area of region in which $z$ lies is sq. units.

Answer 1

$|z-3|\le |z-2|\Rightarrow |z-3{|}^2\le |z-2{|}^2$
$\Rightarrow (z-3)(\overline{z}-3)\le (z-2)(\overline{z}-2)$
$\Rightarrow z+\overline{z}\ge 5$
Assuming $z=x+iy$,
$2x\ge 5\Rightarrow x\ge \frac{5}{2}$
Now,
$|z-3|\le |z-6|\Rightarrow |z-3{|}^2\le |z-6{|}^2$
$\Rightarrow (z-3)(\overline{z}-3)\le (z-6)(\overline{z}-6)$
$\Rightarrow 3z+3\overline{z}\le 27$
$\Rightarrow z+\overline{z}\le \frac{27}{3}$
$\Rightarrow 2x\le 9\Rightarrow x\le \frac{9}{2}$
Therefore, we get
$\frac{5}{2}\le x\le \frac{9}{2}$

Similarly, $0\le y\le 3$
This region is a rectangle with area $A=(\frac{9}{2}-\frac{5}{2})\times (3-0)$
$A=6$ sq.units