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

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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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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