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Example 16 - Convert z = (i - 1)/ cos pi/3 + i sin pi/3 - Polar representation

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Example 16 Convert the complex number z = (𝑖 βˆ’ 1)/γ€–cos 〗⁑〖π/3 + 𝑖 sin⁑〖 Ο€/3γ€— γ€— in the polar form. Let z = (𝑖 βˆ’ 1)/cos⁑〖 Ο€/3 + 𝑖 sin⁑〖 Ο€/3γ€— γ€— = (𝑖 βˆ’ 1)/(cos⁑〖( 180/3 ) + 𝑖 γ€–sin ( 〗⁑〖180/3γ€— γ€— ) ) = (𝑖 βˆ’1)/(cos⁑〖60Β° + 𝑖 sin⁑〖60Β°γ€— γ€— ) = (𝑖 βˆ’1)/(1/2 + √3/2 𝑖) = (𝑖 βˆ’ 1)/( (1 + √3 𝑖)/2) = (2 ( 𝑖 βˆ’1 ))/( 1 + √3 𝑖) Rationalizing = (2 ( 𝑖 βˆ’1))/(1+ √3 𝑖) Γ— (1 βˆ’ √3 𝑖)/(1 βˆ’ √3 𝑖) = (2 (1 βˆ’ 𝑖) (1 βˆ’ √3 𝑖))/((1+ √3 𝑖) (1 βˆ’ √3 𝑖)) = (2 [𝑖 (1 βˆ’ √3 𝑖) βˆ’1 (1 βˆ’ √3 𝑖)])/((1+ √3 𝑖) (1 βˆ’ √3 𝑖)) = (2[𝑖 βˆ’ √3 𝑖2 βˆ’ 1 +√3 𝑖])/((1 βˆ’ √3 𝑖) (1 βˆ’ √3 𝑖)) Using (a – b) (a + b) = a2 – b2 = (2[ βˆ’1 + 𝑖 + √3 𝑖 βˆ’ √3 𝑖2])/(1^2 βˆ’(√3 𝑖)^2 ) = (2 [βˆ’1 + 𝑖 + √3 𝑖 βˆ’ √3 𝑖 (𝑖2)])/(1 βˆ’ 3𝑖2) Putting 𝑖2 = - 1 = (2 [βˆ’1 + 𝑖 + √3 𝑖 βˆ’ √3 𝑖 (βˆ’1 )])/(1 βˆ’ 3(βˆ’1)) = (2 [βˆ’1 + 𝑖 + √3 𝑖+ √3])/(1 + 3) = (2 [ ( βˆ’1 + √3 ) + ( 𝑖+ √3 𝑖 ) ])/4 = (( βˆ’1 + √3 )+𝑖 ( 1+ √3 ) )/2 = (√3 βˆ’1)/2 + 𝑖 (√3 + 1)/2 Now z = (√3 βˆ’1)/2 + 𝑖 (√3 + 1)/2 Now z = (√3 βˆ’1)/2 + 𝑖 (√3 + 1)/2 Let Polar form be z = r ( cos ΞΈ + 𝑖 sin ΞΈ ) From (1) and (2) (√3 βˆ’1)/2 + 𝑖 (√3 + 1)/2 = r ( cos ΞΈ + 𝑖 sin ΞΈ ) (√3 βˆ’1)/2 + 𝑖 (√3 + 1)/2 = r cos ΞΈ + 𝑖r sin ΞΈ Adding (3) + (4) (4 βˆ’ 2√3 )/4 + (4 + 2√3 )/4 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 1/4 ( 4 - 2√3 + 4 + 2√3 ) = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 1/4 ( 4 + 4 - 2√3 + 2√3 ) = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 1/4 ( 8 – 0 ) = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 8/4 = r2 Γ— 1 2 = r2 √2 = r r = √2 Now finding argument (√3 βˆ’1)/2 + 𝑖 (√3 + 1)/2 = r cos ΞΈ + 𝑖r sin ΞΈ Comparing real part (√3 βˆ’1)/2 = r cos ΞΈ Put r = √2 (√3 βˆ’1)/2 = √2 cos ΞΈ (√3 βˆ’1)/(2√2) = cos ΞΈ Hence, cos ΞΈ = (√3 βˆ’1)/(2√2) & sin ΞΈ = (√3 + 1)/(2√2) Hence, cos ΞΈ = (√3 βˆ’1)/(2√2) & sin ΞΈ = (√3 + 1)/(2√2) Since sin ΞΈ and cos ΞΈ both Positive Hence ΞΈ lies in the lst Quadrant Argument (ΞΈ ) of z = 75o = 75o Γ— πœ‹/180 = 5πœ‹/12 Hence Polar form of z = r ( cos ΞΈ + i sin ΞΈ ) = √2 ( cos 5πœ‹/12 + i sin 5πœ‹/12) Hence, Argument = 75Β° = 75 Γ— πœ‹/180 = 5πœ‹/12 Hence , r = √2 , & ΞΈ = 5πœ‹/12 Thus, Polar form of z = r ( cos ΞΈ + i sin ΞΈ ) = √2 ( cos 5πœ‹/12 + i sin 5πœ‹/12)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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