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 :

Express the complex number - 1 + 3 i in polar form and exponential form.

Given complex number is - 1 + 3 i . Let, z = - 1 + i 3 = a + i b Here, a = - 1 , b = 3 and z = - 1 + i 3 belongs to the second quadrant. To find the polar form and exponential form of a complex, we need to find the modulus and argument of the complex number. ∴ Modulus of z = z = - 1 2 + ( 3 ) 2 ⇒ z = 4 ⇒ z = 2 Argument θ of z = - 1 + i 3 is given by θ = π - tan - 1 3 - 1 ∵ θ = π - tan - 1 Im z Re z , when z belongs to 2 nd quadrant ⇒ θ = π - tan - 1 ( 3 ) ⇒ θ = π - π 3 = 2 π 3 Now, polar form a complex number z is defined as z cos θ + i sin θ ∴ Polar form of z = 2 cos 2 π 3 + i sin 2 π 3 Exponential form of complex number is defined z = z e i θ , where z is the modulus of the complex number and θ is the argument of z . ∴ Exponential form of z = 2 e i 2 π 3 Hence polar and exponential form of the given complex number is 2 cos 2 π 3 + i sin 2 π 3 and 2 e i 2 π 3 respectively.