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

Example 16 - Chapter 5 Class 11 Complex Numbers - Part 2
Example 16 - Chapter 5 Class 11 Complex Numbers - Part 3 Example 16 - Chapter 5 Class 11 Complex Numbers - Part 4 Example 16 - Chapter 5 Class 11 Complex Numbers - Part 5 Example 16 - Chapter 5 Class 11 Complex Numbers - Part 6 Example 16 - Chapter 5 Class 11 Complex Numbers - Part 7 Example 16 - Chapter 5 Class 11 Complex Numbers - Part 8 Example 16 - Chapter 5 Class 11 Complex Numbers - Part 9

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.