Let Z=x+iy,∀x,y∈R A={z:|z|≤2} and B={z:(z+2y)+z≥4} ∴|z|≤2 and z+2y+z≥4 √x2+y2≤2 and x+iy+2y+x−iy≥4 x2+y2≤4 and 2x+2y≥4 ⇒x+y≥2
Point of intersection x2+y2=4 and x+y=2 ⇒y=2−x x2+(2−x)2=4 x2+4−4x+x2=4 2x2−4x=0 ⇒2x(x−2)=0 x=0,x=2 ⇒y=2,y=0 ∴ A‌⋂B= Shaded Area