Draw a set of values

Question

I have no idea how to draw properly those inequalities:
a) $\left | \frac {z+3}{z-2i} \right | \geqslant 1$
b) $\left | z^{2}+4 \right |\leqslant \left |z-2i \right |$
While trying to solwe a) I got $y\geqslant-\frac{3}{2}x-\frac{5}{4}$, but it seems not to be right solution.

Answer 1

$|z+3|^2 = |x+3 + yi|^2 = (x+3)^2 + y^2$, and $|z-2i|^2 = |x + (y-2)i|^2 = x^2 + (y-2)^2$. Thus the 1st inequality translates into: $(x+3)^2 + y^2 \geq x^2 + (y-2)^2 \iff 6x + 4y \geq -5$. The graph of this is a half-plane that contains the origin whose boundary is the line: $6x+4y + 5 = 0$.

But note that we must have: $z \neq 2i$ or $x + yi \neq 0 + 2i$. Thus either $x = 0, y \neq 2$ or $x \neq 0, y = 2$ or $x \neq 0, y \neq 2$. Thus the graph of a) is the half-plane minus the set of points $A = \{z: z =2i\}$.

Answer 2

For (a), $|\frac{z+3}{z-2i}| \geq 1 \implies \frac{|z+3|}{|z-2i|} \geq 1 \\ |z+3| \geq |z-2i|$.

We can geometrically interpret this as: the distance of z from $(-3,0)$ is greater than or equal to its distance from $(0,2)$. We thus then draw the perpendicular bisector of the line segment, and shade in the appropriate line segment (it should be obvious to you once you draw the diagram).

As for (b), $|z^2+4| \leq |z-2i| \implies |z-2i||z+2i| \leq |z-2i| \\ |z+2i| \leq 1$

Geometrically, this is a circle centre $(0,-2)$, radius 1 unit.