If z=(λ+3)−i5−λ2, then the locus of z is

Let z = x + iy. Then,x=λ+3 and y=−5−λ2⇒ (x−3)2=λ2         (1)and y2=5−λ2             (2)From Eqs. (1) and (2),(x−3)2=5−y2 or (x−3)2+y2=5Obviously it is a semicircle as y < 0. Hence, part of the circle lies below the x-axis.