Miscellaneous

Chapter 5 Class 11 Complex Numbers
Serial order wise

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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))

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.