Express the complex number 1 + i 3 in modulus-amplitude form.

Let, 1 + i 3 = r cos θ + r i sin θ Comparing real and imaginary parts, we get r cos θ = 1 ...(i) and r sin θ = 3 ...(ii) ⇒ r = 1 + i 3 = 1 + 3 2 = 1 + 3 = 4 = 2 Now, ( i i ) ( i ) = sin θ cos θ = tan θ = 3 ⇒ θ = π 3 ∴ 1 + i 3 = 2 cos π 3 + i sin π 3

Express the complex number - 3 + i in modulus-amplitude form.

Let, - 3 + i = r cos θ + r i sin θ Comparing real and imaginary parts r cos θ = - 3 ...(i) and r sin θ = 1 ...(ii) r is the modulus of the given complex number. ⇒ r = | i - 3 | = 1 + ( - 3 ) 2 = 1 + 3 = 4 = 2 Then, sin θ cos θ = tan θ = - 1 3 ⇒ θ = 5 π 6 ∴ - 3 + i = 2 cos 5 π 6 + isin 5 π 6

Express the complex number - 1 - i 3 in modulus-amplitude form.

Let, - 1 - i 3 = r cos θ + r i sin θ Comparing real and imaginary parts, we get r cos θ = - 1 ...(i) and r sin θ = - 3 ...(ii) Here, r is the modulus of the given complex number ⇒ r = | - 1 - i 3 | = 1 + ( - 3 ) 2 = 1 + 3 = 4 = 2 Now, ( i i ) ( i ) = tan θ = 3 ⇒ θ = 4 π 3 , since - 1 - i 3 = - 1 , - 3 is in III quadrant. ∴ - 1 - i 3 = 2 cos 4 π 3 + i sin 4 π 3