For \(z=x+iy\), the first condition becomes \((x-1)^2+(y+2)^2\leq x^2+y^2\), or \(y\leq x/2-5/4\). Its boundary is the perpendicular bisector of the segment from \(0\) to \(1-2i\). The second condition is the closed disk centred at \((2,0)\), radius 1. Shade the part of this disk below the straight line; both boundaries are included.