Example 16 - Chapter 5 Class 11 Complex Numbers (Term 1)

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Example 16 Convert the complex number z = (𝑖 − 1)/〖cos 〗⁡〖π/3 + 𝑖 sin⁡〖 π/3〗 〗 in the polar form. Let z = (𝑖 − 1)/cos⁡〖 π/3 + 𝑖 sin⁡〖 π/3〗 〗 = (𝑖 − 1)/(cos⁡〖( 180/3 ) + 𝑖 〖sin ( 〗⁡〖180/3〗 〗 ) ) = (𝑖 −1)/(cos⁡〖60° + 𝑖 sin⁡〖60°〗 〗 ) = (𝑖 −1)/(1/2 + √3/2 𝑖) = (𝑖 − 1)/( (1 + √3 𝑖)/2) = (2 ( 𝑖 −1 ))/( 1 + √3 𝑖) Rationalizing = (2 ( 𝑖 −1))/(1+ √3 𝑖) × (1 − √3 𝑖)/(1 − √3 𝑖) = (2 (1 − 𝑖) (1 − √3 𝑖))/((1+ √3 𝑖) (1 − √3 𝑖)) = (2 [𝑖 (1 − √3 𝑖) −1 (1 − √3 𝑖)])/((1+ √3 𝑖) (1 − √3 𝑖)) = (2[𝑖 − √3 𝑖2 − 1 +√3 𝑖])/((1 − √3 𝑖) (1 − √3 𝑖)) Using (a – b) (a + b) = a2 – b2 = (2[ −1 + 𝑖 + √3 𝑖 − √3 𝑖2])/(1^2 −(√3 𝑖)^2 ) = (2 [−1 + 𝑖 + √3 𝑖 − √3 𝑖 (𝑖2)])/(1 − 3𝑖2) Putting 𝑖2 = - 1 = (2 [−1 + 𝑖 + √3 𝑖 − √3 𝑖 (−1 )])/(1 − 3(−1)) = (2 [−1 + 𝑖 + √3 𝑖+ √3])/(1 + 3) = (2 [ ( −1 + √3 ) + ( 𝑖+ √3 𝑖 ) ])/4 = (( −1 + √3 )+𝑖 ( 1+ √3 ) )/2 = (√3 −1)/2 + 𝑖 (√3 + 1)/2 Now z = (√3 −1)/2 + 𝑖 (√3 + 1)/2 Now z = (√3 −1)/2 + 𝑖 (√3 + 1)/2 Let Polar form be z = r ( cos θ + 𝑖 sin θ ) From (1) and (2) (√3 −1)/2 + 𝑖 (√3 + 1)/2 = r ( cos θ + 𝑖 sin θ ) (√3 −1)/2 + 𝑖 (√3 + 1)/2 = r cos θ + 𝑖r sin θ Adding (3) + (4) (4 − 2√3 )/4 + (4 + 2√3 )/4 = r2 cos2 θ + r2 sin2 θ 1/4 ( 4 - 2√3 + 4 + 2√3 ) = r2 ( cos2 θ + sin2 θ ) 1/4 ( 4 + 4 - 2√3 + 2√3 ) = r2 ( cos2 θ + sin2 θ ) 1/4 ( 8 – 0 ) = r2 ( cos2 θ + sin2 θ ) 8/4 = r2 × 1 2 = r2 √2 = r r = √2 Now finding argument (√3 −1)/2 + 𝑖 (√3 + 1)/2 = r cos θ + 𝑖r sin θ Comparing real part (√3 −1)/2 = r cos θ Put r = √2 (√3 −1)/2 = √2 cos θ (√3 −1)/(2√2) = cos θ Hence, cos θ = (√3 −1)/(2√2) & sin θ = (√3 + 1)/(2√2) Hence, cos θ = (√3 −1)/(2√2) & sin θ = (√3 + 1)/(2√2) Since sin θ and cos θ both Positive Hence θ lies in the lst Quadrant Argument (θ ) of z = 75o = 75o × 𝜋/180 = 5𝜋/12 Hence Polar form of z = r ( cos θ + i sin θ ) = √2 ( cos 5𝜋/12 + i sin 5𝜋/12) Hence, Argument = 75° = 75 × 𝜋/180 = 5𝜋/12 Hence , r = √2 , & θ = 5𝜋/12 Thus, Polar form of z = r ( cos θ + i sin θ ) = √2 ( cos 5𝜋/12 + i sin 5𝜋/12)