Let z=x+iy,zˉ=x−iy ⇒⇒⇒ Now, z=1−zˉx+iy=1−(x−iy)2x=1⇒x=21 Now, ∣z∣=1⇒x2+y2=1⇒y2=1−x2y=±23 Now, tanθ=xy(θ is the argument ) =23÷21 (+ve since only principal argument) =3⇒θ=tan−13=3π Hence, z is not a real number So, statement-1 is false and 2 is true.