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 sq. units in which $z$ lies is

• A. $9$
• B. $6$
• C. $2$
• D. $14$

Answer 1

The correct option is B $6$

$|z-3|\le |z-2|$
$\Rightarrow |z-3{|}^2\le |z-2{|}^2$
$\Rightarrow (z-3)(\overline{z}-3)\le (z-2)(\overline{z}-2)$
$\Rightarrow z+\overline{z}\ge 5$
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, $-3\le y\le 0$
This region is a rectangle with area $A=|\frac{9}{2}-\frac{5}{2}|\times |0-(-3)|$
$A=6$ sq. units