Given : |z| - z =1+2i If z = x + iy, then this equation reduces to | x + iy | - (x + iy) = 1 + 2i ⇒ (√x2+y2−x)+(−iy)=1+2i On comparing real and imaginary parts of both sides of this equation, we get √x2+y2−x=1 ⇒ √x2+y2=1+x ⇒x2+y2=(1+x)2 ⇒x2+y2=1+x2+2x ⇒y2=1+2x...(i) and -y = 2 ⇒ y = -2 Putting this value in eq.(i), we get (−2)2=1+2x ⇒ 2x=3⇒x=