Question

Find the points of the complex plane which satisfy the following inequalities.
$|Z-4|<1.$

Answer 1

Let $z=x+yi$

$|z-4|=|x+yi-4|$

$=$ $|(x-4)+yi|$

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

$=\sqrt{x^2+16-8x+y^2}$

$\therefore$ $|z-4|<1$

$\Rightarrow \sqrt{x^2+y^2-8x+16}<1$

$\Rightarrow x^2+y^2-8x+16<1$

$\Rightarrow x^2+y^2-8x+15<0$

The center of the circle $x^2+y^2-8x+15=0$ is $(4,0)$ and radius equal to $\sqrt{16+0+15}=1$

Here the set of points satisfying $|z-4|<1$ consists of all points in interior of circle.