1. Class 11
2. Important Question for exams Class 11

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 + π2β 4π) Putting π2 = -1 = (1 + 7π)/(4 + ( β1 ) 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 β4π) (3 β4π)) = (1 (3+4π) +7π (3 + 4π ))/((3 β4π) (3 + 4π)) = (3 + 4π + 21π + 28π2)/((3 β 4π) (3 + 4π)) = (3 + 25π + 28π2)/((3 β 4π) (3 + 4π)) Using (a β b) (a + b) = a2 β b2 = (3 + 25π + 28π2)/((3)2β (4π)2) = (3 + 25π + 28π2)/(9 β 16π2) Putting π2 = 1 = (3 + 25π + 28 (β1 ))/(9 β16 ( β1 )) = (3 + 25π β 28)/(9 + 16) = (3 β 28 + 25π)/25 = (β 25 + 25π)/25 = ( 25 ( β1 + π ))/25 = - 1 + π Hence, z = β 1 + π Let polar form be z = π (cosβ‘ΞΈ+πsinβ‘ΞΈ ) From (1) and (2) β1 + π = r (cos ΞΈ + π sin ΞΈ) β1 + 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 = π2 cos2ΞΈ Adding (3) and (4) 1 + 1 = π2 cos2 ΞΈ + π2 sin2 ΞΈ 1 + 1 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 2 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 2 = r2 Γ 1 2 = r2 β2 = r r = β2 Now finding argument β1 + π = r cos ΞΈ + π r sin ΞΈ Comparing real part β 1 = r cos ΞΈ Putting r =β2 β 1 = β2 cos ΞΈ (β 1 )/β2 = cos ΞΈ cos ΞΈ = (β 1 )/β2 Hence, cos ΞΈ = (β 1 )/β2 & sin ΞΈ = ( 1)/β2 Hence, cos ΞΈ = (β 1 )/β2 & sin ΞΈ = ( 1)/β2 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 = β2 and ΞΈ = 3π/4 Polar form of π§=π (cosβ‘ΞΈ+sinβ‘ΞΈ ) = β2 (cos(3π/4)+sin(3π/4)) Misc 5 Convert the following in the polar form: (ii) ( 1 + 3π)/(1 β 2π) Let z = ( 1 + 3π)/(1 β 2π) Rationalizing = (1 + 3π)/(1 β 2π) Γ (1 + 2π)/(1 + 2π) = ((1 + 3π) (1 + 2π ))/((1 β 2π) (1 + 2π)) = (1 (1 + 2π) + 3π (1 + 2π ))/((1 β 2π) (1 + 2π)) = (1 +2π +3π +6π2)/((1 β 2π) (1 + 2π)) = (1 + 5π +6π2)/((1 β2π) (1 +2π)) Using (a β b) (a + b) = a2 β b2 = (1 + 5π +6π2)/((1)2β (2π)2) = (1 + 5π +6π2)/(1 β4π2) Putting π2 = β1 = (1 + 5π +6 (β1 ))/(1 β4 ( β1 )) = (1 + 5π β6)/(1 +4) = (1 β6 + 5π)/5 = (β 5 + 5π)/5 = (β 5 ( β1 + π ))/5 = - 1 + π Hence, z = β 1 + π Let polar form be z = π (cosβ‘ΞΈ+πsinβ‘ΞΈ ) From (1) and (2) β1 + π = r (cos ΞΈ + π sin ΞΈ) β1 + 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 = π2 cos2ΞΈ Adding (3) and (4) 1 + 1 = π2 cos2 ΞΈ + π2 sin2 ΞΈ 1 + 1 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 2 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 2 = r2 Γ 1 2 = r2 β2 = r r = β2 Now finding argument β1 + π = r cos ΞΈ + π r sin ΞΈ Comparing real part β 1 = r cos ΞΈ Putting r =β2 β 1 = β2 cos ΞΈ (β 1 )/β2 = cos ΞΈ cos ΞΈ = (β 1 )/β2 Hence, cos ΞΈ = (β 1 )/β2 & sin ΞΈ = ( 1)/β2 Hence, cos ΞΈ = (β 1 )/β2 & sin ΞΈ = ( 1)/β2 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 = β2 and ΞΈ = 3π/4 Polar form of π§=π (cosβ‘ΞΈ+sinβ‘ΞΈ ) = β2 (cos(3π/4)+sin(3π/4))

Class 11
Important Question for exams Class 11

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.