Question

Find the points of the plane which satisfy the following inequalities.
$|z+i-2|⩽2$

Answer 1

Let $z=x+iy$

Then, $|z+i-2|\le 2$

$\Rightarrow |x+iy+i-2|\le 2$

$\Rightarrow |(x-2)+i(y+1)|\le 2$

$\Rightarrow \sqrt{{(x-2)}^2+{(y+1)}^2}\le 2$

$\Rightarrow {(x-2)}^2+{(y+1)}^2\le 4$

$\Rightarrow x^2+4-4x+y^2+1+2y\le 4$

$\Rightarrow x^2+y^2-4x+2y+1\le 4-4$

$\Rightarrow x^2+y^2-4x+2y+1\le 0$

Now $x^2+y^2-4x+2y+1=0$ is a circle with center $(2,-1)$ and radius equal to $\sqrt{4+1-1}=2$

The above circle cuts $x$ axis where $y=0$. Then we get

$x^2-4x+1=0$

$x=\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm 2\sqrt{3}}{2}=2\pm \sqrt{3}$

$\Rightarrow x=2\pm 1.732=3.732$

The circle cuts $y$ axis at where $x=0$

$y^2+2y+1$

$\Rightarrow {(y+1)}^2=0$

$\Rightarrow y=(-1,-1)$. Thus the circle touches $y$ axis at $(0,-1)$.