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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

Ex 5.2, 6 Deleted for CBSE Board 2023 Exams

Ex 5.2, 7 Important Deleted for CBSE Board 2023 Exams

Ex 5.2, 8 Deleted for CBSE Board 2023 Exams

Last updated at Sept. 3, 2021 by Teachoo

Ex5.2, 3 Convert the given complex number in polar form: 1 โ i Given ๐ง = 1 โ ๐ Let polar form be z = ๐ (cosโกฮธ+๐ sinโกฮธ ) From (1) and (2) 1 - ๐ = r (cos ฮธ + ๐ sin ฮธ) 1 โ ๐ = r cos ฮธ + ๐ r sin ฮธ Comparing real part 1 = r cos ฮธ Squaring both sides (1)2 =( ๐ cosโกฮธ )^2 1 = ๐2 cos2ฮธ Adding (3) and (4) 1 + 1 = ๐2 cos2 ฮธ + ๐2 sin2 ฮธ 1 + 1 = r2 cos2 ฮธ + r2 sin2 ฮธ 2 = r2 ( cos2 ฮธ + sin2 ฮธ ) 2 = r2 ร 1 2 = r2 โ2 = r r = โ2 Now finding argument 1 โ ๐ = r cos ฮธ + ๐ r sin ฮธ Comparing real part 1 = r cos ฮธ Putting r =โ2 1 = โ2 cos ฮธ 1/โ2 = cos ฮธ cos ฮธ = 1/โ2 Hence, cos ฮธ = 1/โ2 & sin ฮธ = (โ 1)/โ2 Hence, cos ฮธ = 1/โ2 & sin ฮธ = (โ 1)/โ2 Since, sin ฮธ is negative and cos ฮธ is positive, Hence, ฮธ lies in IVth quadrant Argument = โ 45ยฐ = โ 45ยฐ ร ๐/(180ยฐ) = (โ ๐)/4 Hence, argument of ๐ง = (โ ๐)/4 Hence ๐ = โ2 and ฮธ = ( โ ๐)/4 Polar form of ๐ง=๐ (cosโกฮธ+sinโกฮธ ) = โ2 (cos(( โ ๐)/4)+sin(( โ ๐)/4))