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Ex 5.2, 4 - Convert in polar form: -1 + i - Complex number - Ex 5.2

Ex 5.2, 4 - Chapter 5 Class 11 Complex Numbers - Part 2
Ex 5.2, 4 - Chapter 5 Class 11 Complex Numbers - Part 3
Ex 5.2, 4 - Chapter 5 Class 11 Complex Numbers - Part 4
Ex 5.2, 4 - Chapter 5 Class 11 Complex Numbers - Part 5

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Transcript

Ex 5.2, 4 Convert the given complex number in polar form: โ€“ 1 + i Given ๐‘ง = โˆ’1+ ๐‘– Let polar form be ใ€–๐‘ง = ๐‘Ÿ (cosใ€—โกฮธ+๐‘– sinโกฮธ) From (1) & (2) โˆ’ 1+ ๐‘– = r ( cosโกฮธ + ๐‘– sinโกฮธ) โˆ’ 1+ ๐‘– = rใ€– cosใ€—โกฮธ + ๐‘– r sinโกฮธ Adding ( 3 ) and ( 4 ) 1 + 1 = ๐‘Ÿ2 cos2 ฮธ+ ๐‘Ÿ2 sin2ฮธ 2 = ๐‘Ÿ2 ( cos2 ฮธ+ sin2 ฮธ) 2 = ๐‘Ÿ2 ร— 1 2 = ๐‘Ÿ2 โˆš2 = ๐‘Ÿ ๐‘Ÿ = โˆš2 Finding argument โˆ’ 1+ ๐‘– = rใ€– cosใ€—โกฮธ + ๐‘– r sinโกฮธ Hence, sin ฮธ = 1/โˆš2 & cos ฮธ = (โˆ’ 1)/โˆš2 Hence, sin ฮธ = 1/โˆš2 & cos ฮธ = (โˆ’ 1)/โˆš2 Here, sin ฮธ is positive and cos ฮธ is negative, Hence, ฮธ lies in IInd quadrant Argument = 180ยฐ โ€“ 45ยฐ = 135ยฐ = 135ยฐ ร— ๐œ‹/180o = ( 3 ๐œ‹)/4 So argument of z = ( 3 ๐œ‹)/4 Hence ๐‘Ÿ = โˆš2 and ฮธ = 3๐œ‹/4 Polar form of z = r (cos ฮธ + sin ฮธ) = โˆš2 (cos (( 3 ๐œ‹)/4)+ ๐‘– sin(( 3 ๐œ‹)/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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.