⇒Re(z2−i)=2 Put z=x+iy Let's find the value of z2−i ⇒(x+iy)2−i ⇒x2+2xyi−y2−i.......[(a+b)2=a2+2ab+b2] Now take real part and imaginary part collectively ⇒(x2−y2)+i(2xy−1) So the real part(z2−i) ⇒Re(z2−i)=x2−y2=2 ⇒x2−y2=2 So it is a rectangular hyperbola.