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Ex 5.2

Ex 5.2, 1

Ex 5.2, 2 Important

Ex 5.2, 3 Important Deleted for CBSE Board 2023 Exams You are here

Ex 5.2, 4 Deleted for CBSE Board 2023 Exams

Ex 5.2, 5 Important Deleted for CBSE Board 2023 Exams

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Ex 5.2, 7 Important Deleted for CBSE Board 2023 Exams

Ex 5.2, 8 Deleted for CBSE Board 2023 Exams

Last updated at March 22, 2023 by Teachoo

Ex5.2, 3 Convert the given complex number in polar form: 1 β i Given π§ = 1 β π Let polar form be z = π (cosβ‘ΞΈ+π sinβ‘ΞΈ ) From (1) and (2) 1 - π = r (cos ΞΈ + π sin ΞΈ) 1 β π = r cos ΞΈ + π r sin ΞΈ Comparing real part 1 = r cos ΞΈ Squaring both sides (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 1 β π = r cos ΞΈ + π r sin ΞΈ Comparing real part 1 = r cos ΞΈ Putting r =β2 1 = β2 cos ΞΈ 1/β2 = cos ΞΈ cos ΞΈ = 1/β2 Hence, cos ΞΈ = 1/β2 & sin ΞΈ = (β 1)/β2 Hence, cos ΞΈ = 1/β2 & sin ΞΈ = (β 1)/β2 Since, sin ΞΈ is negative and cos ΞΈ is positive, Hence, ΞΈ lies in IVth quadrant Argument = β 45Β° = β 45Β° Γ π/(180Β°) = (β π)/4 Hence, argument of π§ = (β π)/4 Hence π = β2 and ΞΈ = ( β π)/4 Polar form of π§=π (cosβ‘ΞΈ+sinβ‘ΞΈ ) = β2 (cos(( β π)/4)+sin(( β π)/4))