Last updated at May 29, 2018 by Teachoo

Transcript

Example 16 Convert the complex number z = (π β 1)/γcos γβ‘γΟ/3 + π sinβ‘γ Ο/3γ γ in the polar form. Let z = (π β 1)/cosβ‘γ Ο/3 + π sinβ‘γ Ο/3γ γ = (π β 1)/(cosβ‘γ( 180/3 ) + π γsin ( γβ‘γ180/3γ γ ) ) = (π β1)/(cosβ‘γ60Β° + π sinβ‘γ60Β°γ γ ) = (π β1)/(1/2 + β3/2 π) = (π β 1)/( (1 + β3 π)/2) = (2 ( π β1 ))/( 1 + β3 π) Rationalizing = (2 ( π β1))/(1+ β3 π) Γ (1 β β3 π)/(1 β β3 π) = (2 (1 β π) (1 β β3 π))/((1+ β3 π) (1 β β3 π)) = (2 [π (1 β β3 π) β1 (1 β β3 π)])/((1+ β3 π) (1 β β3 π)) = (2[π β β3 π2 β 1 +β3 π])/((1 β β3 π) (1 β β3 π)) Using (a β b) (a + b) = a2 β b2 = (2[ β1 + π + β3 π β β3 π2])/(1^2 β(β3 π)^2 ) = (2 [β1 + π + β3 π β β3 π (π2)])/(1 β 3π2) Putting π2 = - 1 = (2 [β1 + π + β3 π β β3 π (β1 )])/(1 β 3(β1)) = (2 [β1 + π + β3 π+ β3])/(1 + 3) = (2 [ ( β1 + β3 ) + ( π+ β3 π ) ])/4 = (( β1 + β3 )+π ( 1+ β3 ) )/2 = (β3 β1)/2 + π (β3 + 1)/2 Now z = (β3 β1)/2 + π (β3 + 1)/2 Now z = (β3 β1)/2 + π (β3 + 1)/2 Let Polar form be z = r ( cos ΞΈ + π sin ΞΈ ) From (1) and (2) (β3 β1)/2 + π (β3 + 1)/2 = r ( cos ΞΈ + π sin ΞΈ ) (β3 β1)/2 + π (β3 + 1)/2 = r cos ΞΈ + πr sin ΞΈ Adding (3) + (4) (4 β 2β3 )/4 + (4 + 2β3 )/4 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 1/4 ( 4 - 2β3 + 4 + 2β3 ) = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 1/4 ( 4 + 4 - 2β3 + 2β3 ) = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 1/4 ( 8 β 0 ) = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 8/4 = r2 Γ 1 2 = r2 β2 = r r = β2 Now finding argument (β3 β1)/2 + π (β3 + 1)/2 = r cos ΞΈ + πr sin ΞΈ Comparing real part (β3 β1)/2 = r cos ΞΈ Put r = β2 (β3 β1)/2 = β2 cos ΞΈ (β3 β1)/(2β2) = cos ΞΈ Hence, cos ΞΈ = (β3 β1)/(2β2) & sin ΞΈ = (β3 + 1)/(2β2) Hence, cos ΞΈ = (β3 β1)/(2β2) & sin ΞΈ = (β3 + 1)/(2β2) Since sin ΞΈ and cos ΞΈ both Positive Hence ΞΈ lies in the lst Quadrant Argument (ΞΈ ) of z = 75o = 75o Γ π/180 = 5π/12 Hence Polar form of z = r ( cos ΞΈ + i sin ΞΈ ) = β2 ( cos 5π/12 + i sin 5π/12) Hence, Argument = 75Β° = 75 Γ π/180 = 5π/12 Hence , r = β2 , & ΞΈ = 5π/12 Thus, Polar form of z = r ( cos ΞΈ + i sin ΞΈ ) = β2 ( cos 5π/12 + i sin 5π/12)

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.