Given that ∣zω∣=1⇒∣z∣∣ω∣=1⇒∣z∣=∣ω∣1 and Arg(z)−Arg(ω)=2π⇒Arg(ωz)=2π When argument of a complex number is 2π, it means it is making an angle of 2π with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis. So, ωz is a purely imaginary number that means there is no real part in this complex number. So we can assume, ωz=ki⇒ωz=∣ki∣⇒ωz=∣k∣∣i∣⇒ωz=k [as ∣i∣=1]⇒∣z∣×∣ω∣1=k⇒∣z∣×∣z∣=k[. as ∣ω∣1=∣z∣]⇒∣z∣2=k⇒∣z∣=k∴∣ω∣=k1 As ωz is imaginary so we can write, ωz=−ωz [When z is imaginary then z=−z ] ⇒zω=−zω⇒zω=−ωz⋅ω⋅ω⇒zω=−ωz⋅∣ω∣2⇒zω=−ki⋅(k1)2⇒zω=−ki⋅k1⇒zω=−i