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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

Ex 5.2, 4 Deleted for CBSE Board 2023 Exams You are here

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 Dec. 8, 2016 by Teachoo

Ex 5.2, 4 Convert the given complex number in polar form: โ 1 + i Given ๐ง = โ1+ ๐ Let polar form be ใ๐ง = ๐ (cosใโกฮธ+๐ sinโกฮธ) From (1) & (2) โ 1+ ๐ = r ( cosโกฮธ + ๐ sinโกฮธ) โ 1+ ๐ = rใ cosใโกฮธ + ๐ r sinโกฮธ Adding ( 3 ) and ( 4 ) 1 + 1 = ๐2 cos2 ฮธ+ ๐2 sin2ฮธ 2 = ๐2 ( cos2 ฮธ+ sin2 ฮธ) 2 = ๐2 ร 1 2 = ๐2 โ2 = ๐ ๐ = โ2 Finding argument โ 1+ ๐ = rใ cosใโกฮธ + ๐ r sinโกฮธ Hence, sin ฮธ = 1/โ2 & cos ฮธ = (โ 1)/โ2 Hence, sin ฮธ = 1/โ2 & cos ฮธ = (โ 1)/โ2 Here, sin ฮธ is positive and cos ฮธ is negative, Hence, ฮธ lies in IInd quadrant Argument = 180ยฐ โ 45ยฐ = 135ยฐ = 135ยฐ ร ๐/180o = ( 3 ๐)/4 So argument of z = ( 3 ๐)/4 Hence ๐ = โ2 and ฮธ = 3๐/4 Polar form of z = r (cos ฮธ + sin ฮธ) = โ2 (cos (( 3 ๐)/4)+ ๐ sin(( 3 ๐)/4))