Sketch complex inequalities

Question

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.

Answer 1

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.

Answer 2

For the first one by $z=x+iy$ we have

$$|z-2i|<1 \iff |z-2i|^2<1\iff x^2+(y-2)^2<1$$

and for the second one

$$|z-1-2i|>3\iff |z-1-2i|^2>9\iff (x-1)^2+(y-2)^2>9$$

Answer 3

@Tuki, your intuition is excellent, the formula for $|z|$ as well, but you applied it wrongly, $Re (z-2i) \neq z.$

If you prefer to continue on your own, replace $z$ by $x+iy.$