Last updated at Sept. 3, 2021 by Teachoo

Transcript

Misc 5 Convert the following in the polar form: (i) (1 + 7π)/(2 β π)2 Let z = ( 1+7π)/(2 β π)2 Using ( a β b)2 = a2 + b2 β 2ab = ((1 + 7π))/((2)2+ (i)2 β 2 Γ 2 Γ π) = ( 1 + 7π)/(4 + π2β 4π) Putting π2 = -1 = (1 + 7π)/(4 + ( β1 ) β 4π) = (1 + 7π )/(4 β 1 β 4π ) = (1 + 7π)/(3 β 4π) Rationalizing the Same = (1 + 7π)/(3 β 4π) Γ (3 + 4π)/(3 + 4π) = ((1+7π) (3 + 4π ))/((3 β4π) (3 β4π)) = (1 (3+4π) +7π (3 + 4π ))/((3 β4π) (3 + 4π)) = (3 + 4π + 21π + 28π2)/((3 β 4π) (3 + 4π)) = (3 + 25π + 28π2)/((3 β 4π) (3 + 4π)) Using (a β b) (a + b) = a2 β b2 = (3 + 25π + 28π2)/((3)2β (4π)2) = (3 + 25π + 28π2)/(9 β 16π2) Putting π2 = 1 = (3 + 25π + 28 (β1 ))/(9 β16 ( β1 )) = (3 + 25π β 28)/(9 + 16) = (3 β 28 + 25π)/25 = (β 25 + 25π)/25 = ( 25 ( β1 + π ))/25 = - 1 + π Hence, z = β 1 + π Let polar form be z = π (cosβ‘ΞΈ+πsinβ‘ΞΈ ) From (1) and (2) β1 + π = r (cos ΞΈ + π sin ΞΈ) β1 + π = r cos ΞΈ + π r sin ΞΈ Comparing real part β 1 = r cos ΞΈ Squaring both sides (β 1 )2 =( π cosβ‘ΞΈ )^2 1 = π2 cos2ΞΈ Adding (3) and (4) 1 + 1 = π2 cos2 ΞΈ + π2 sin2 ΞΈ 1 + 1 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 2 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 2 = r2 Γ 1 2 = r2 β2 = r r = β2 Now finding argument β1 + π = r cos ΞΈ + π r sin ΞΈ Comparing real part β 1 = r cos ΞΈ Putting r =β2 β 1 = β2 cos ΞΈ (β 1 )/β2 = cos ΞΈ cos ΞΈ = (β 1 )/β2 Hence, cos ΞΈ = (β 1 )/β2 & sin ΞΈ = ( 1)/β2 Hence, cos ΞΈ = (β 1 )/β2 & sin ΞΈ = ( 1)/β2 Since, sin ΞΈ is positive and cos ΞΈ is negative, Hence, ΞΈ lies in IInd quadrant Argument = 180Β° β 45Β° = 135Β° = 135Β° Γ π/(180Β°) = 3π/4 Hence, argument of π§ = 3π/4 Hence π = β2 and ΞΈ = 3π/4 Polar form of π§=π (cosβ‘ΞΈ+sinβ‘ΞΈ ) = β2 (cos(3π/4)+sin(3π/4))

Miscellaneous

Misc 1
Important

Misc 2

Misc 3

Misc 4 Important

Misc 5 (i) Deleted for CBSE Board 2022 Exams You are here

Misc 5 (ii) Important

Misc 6

Misc 7

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11

Misc 12

Misc 13 Important Deleted for CBSE Board 2022 Exams

Misc 14

Misc 15 Important

Misc 16

Misc 17 Important

Misc 18 Important

Misc 19

Misc 20 Important

Chapter 5 Class 11 Complex Numbers (Term 1)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.