How to sketch the region on the complex plane $\{ z: |z-2i| \le |z+6+4i| \}$?

I have a problem of understanding how to find shaded regions in Complex Plane.

\begin{array}{l} |z-2i|\ \geqslant \ |z+6+4i|\\ \\ \sqrt{x^{2} +( y-2)^{2}} =\sqrt{( x+6)^{2} +( y+4)^{2}}\\ x^{2} +( y-2)^{2} =( x+6)^{2} +( y+4)^{2}\\ x^{2} +y^{2} -4y+4=x^{2} +12x+36+y^{2} +8y+16\\ 12x+12y+48=0\\ y=-x-4\\ \end{array}

I can sketch this in complex plane, But I'm confused in how to get shaded regions like this questions.

Solution 1:

Geometrically speaking, the inequality can be understood as a statement about distances: the distance from $z$ to $w_1:=2\mathrm i$ is greater than or equal to the distance from $z$ to $w_2:=-6-4\mathrm i$.

Mark these points on the plane. For a simple start, find all points where the distance to $w_1$ is equal to the distance to $w_2$. Remembering high school geometry (just guessing here, since I don't know the US school system well), those points all lie on the perpendicular bisector between $w_1$ and $w_2$. That's the line you drew.

Now to find the points which are closer to $w_2$ than to $w_1$. You can probably find those yourself now, and shade the region accordingly.

Solution 2:

You are almost there. You need to think what's the geometrical interpretation of $|z-2i| = |z+6+4i|$. You have on the left the distance to the $0+2i$ point in the complex plane, on the right the distance to $-6-4i$ point. Draw the line uniting the two points, then the perpendicular bisector. Any points on this perpendicular are at the same distance from the two points. Now depending on the sign of the inequality, you shade the half plane that contains the point where the distance is smaller. Notice that there is a discrepancy between the title and the first equation. One is $\le$, one is $\ge$.