Find real values of θ for which 4 + 3 i sin θ 1 - 2 i sin θ is purely real.

Let z = 4 + 3 i sin θ 1 - 2 i sin θ be the given complex number z = 4 + 3 i sin θ 1 - 2 i sin θ × 1 + 2 i sin θ 1 + 2 i sin θ ⇒ z = 4 + 8 i sin θ + 3 i sin θ + 6 i 2 sin 2 θ 1 - 4 i 2 sin 2 θ ⇒ z = 4 + ( 11 sin θ ) i - 6 sin 2 θ 1 + 4 sin 2 θ ∵ i 2 = - 1 ⇒ z = 4 - 6 sin 2 θ + ( 11 sin θ ) i 1 + 4 sin 2 θ ∴ Z = 4 - 6 sin 2 θ 1 + 4 sin 2 θ + 11 sin θ 1 + 4 sin 2 θ i Given that z is purely real ⇒ Im ( z ) = 0 ⇒ 11 sin θ 1 + 4 sin 2 θ = 0 ⇒ sin θ = 0 = sin n π , where n ∈ Z ∴ θ = n π , where n ∈ Z Hence, the value of θ for which the given complex number is purely real is θ = n π , where n ∈ Z .

If z = 3 + 5 i , then represent the z , z ¯ , - z , - z ¯ in Argand's diagram.

Given complex number is z = 3 + 5 i Therefore z ¯ = 3 - 5 i , - z = - 3 - 5 i and - z ¯ = - 3 + 5 i The complex numbers z , z ¯ , — z and - z ¯ are represented by the points A ( 3 , 5 ) , B ( 3 , — 5 ) , C ( — 3 , — 5 ) and D ( — 3 , 5 ) respectively in the following figure :