Draw a set of values 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.

$|z+3|^2 = |x+3 + yi|^2 = (x+3)^2 + y^2$, 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: $z \
eq 2i$ or $x + yi \
eq 0 + 2i$. Thus either $x = 0, y \
eq 2$ or $x \
eq 0, y = 2$ or $x \
eq 0, y \
eq 2$. Thus the graph of a) is the half-plane minus the set of points $A = \\{z: z =2i\\}$.