Miscellaneous

Chapter 5 Class 11 Complex Numbers
Serial order wise

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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))