Problem
Sketch complex inequalities to complex plane.
(a)$$|z-2i|<1 $$
(b) $$ |z-1-2i|>3 $$
Attempt to solve
My intuition would tell me that (a) is area inside circle and (b) is area outside of circle in complex plane. However if i try to explicitly solve these inequalities i get answer that doesn't make any sense to me.
I tried with simplified version without inequality ( should contain only points from rim of the circle )
I form expression for length of the radius with Pythagoras.
$$|z|=\sqrt{(\operatorname{Re}(z)^2+(\operatorname{Im}(z))^2}$$
$$ \sqrt{z^2+(-2)^2}=1 $$
$$ z^2+4=1 $$
$$ z^2=-3 $$
$$ z=\pm \sqrt{-3} = \pm i\sqrt{3} $$
I tried to solve where this circles radius is exactly 1. It would seem there is gap in my intuition and cant seem to make sense of this.
Best Answer
For $$|z-2i|<1$$ you are looking for all points in the complex plane whose distance from $2i$ is less that $1$.
That is the open disk centered at $z=2i$ and radius $1$ With center at $(0,2) $and radius $1$ we get $$ x^2+(y-2)^2<1$$
For $$|z-1-2i|>3$$ you are looking for all the points in complex plane whose distance from $ 1+2i$ is greater than $3$.
That is the outside region of the disk with center $(1,2)$ and radius $3$
You can write the inequality to describe it.