Solving $|z-3| \leq|z-1-i|$

Question

I was trying to solve graphicly: $$|z-3| \leq |z-1-i|$$

I plugged x and y in proper places as real componenets of the comlex number yielding in the end $-4x+2y+7 \leq0$

this might be tackled if first eqauted to $0$ hence: $y=2x-3.5$

I need to sketch it also and I suppose that it should be all values which are under the graph of $y=2x-3.5$ on the gauss plane?

Answer 1

For the sketching part: Note that the inequality, viewed geometrically, is satisfied by precisely the points $P$ such that the distance from $P$ to $(3,0)$ is $\le$ the distance from $P$ to $(1,1)$.

Draw the perpendicular bisector $\ell$ of the line segment that joins $(3,0)$ and $(1,1)$. Our points are the points on or "below" line $\ell$.