Example 13 - Find modulus, argument of (1 + i)/(1 - i) - Examples

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Example, 13 Find the modulus and argument of the complex numbers: (i) (1 + ๐‘–)/(1 โˆ’ ๐‘–) , First we solve (1 + ๐‘–)/(1 โˆ’ ๐‘–) Let ๐‘ง = (1 + ๐‘–)/(1 โˆ’ ๐‘–) Rationalizing the same = (1 + ๐‘–)/(1 โˆ’ ๐‘–) ร— (1 + ๐‘–)/(1 + ๐‘–) = (( 1 + ๐‘– ) ( 1 + ๐‘– ))/("(" 1 โˆ’ ๐‘– ) (1 + ๐‘– )) Using (a โ€“ b) (a + b) = a2 โˆ’ b2 = ( 1+ ๐‘– )2/( ( 1 )2 โˆ’ ( ๐‘– )2) Using ( a + b )2 = a2 + b2 + 2ab = ((1)2 + (๐‘–)2 + 2๐‘–)/( (1)2 โˆ’ (๐‘–)2) Putting i2 = โˆ’ 1 = ((1)2 + (โˆ’1) + 2๐‘–)/( 1โˆ’ (โˆ’ 1) ) = (1 โˆ’1 + 2๐‘–)/( 1 + 1) = ( 2๐‘–)/( 2) = ๐‘– = 0 + ๐‘– Hence, ๐‘ง = 0 + ๐‘– Method 1 To calculate modulus of z z = 0 + i Complex number z is of the form x + ๐‘–y Hence x = 0 and y = 1 Modulus of z = โˆš(๐‘ฅ^2+๐‘ฆ2) = โˆš(( 0 )2+(1)2) = โˆš(0+1) = โˆš1 = 1 Modulus of z = 1 Method 2 To calculate modulus of z We have , z = 0 + ๐‘– Let z = r ( cos ฮธ + ๐‘– sin ฮธ ) Here r is modulus, and ฮธ is argument From (1) and (2) 0 + ๐‘– = r ( cos ฮธ + ๐‘– sin ฮธ ) 0 + ๐‘– = r cos ฮธ + ๐‘–r sin ฮธ Comparing real part 0 = r cos ฮธ Squaring both sides (0)2 = ( ๐‘Ÿ cosโกฮธ )2 0 = r2 cos2 ฮธ Adding (3) and (4) 0 + 1 = r2 cos2 ฮธ + r2 sin2 ฮธ 1=๐‘Ÿ2 (cos2 ฮธ+sin2 ฮธ) 1 = r2 (1) 1 = r2 1 = r โ‡’ Modulus of z = 1 Finding argument 0 + ๐‘– = r cos ฮธ + ๐‘–r sin ฮธ Comparing real part 0 = r cos ฮธ Put r = 1 0 = 1 ร— cos ฮธ 0 = cos ฮธ cos ฮธ = 0 Hence, cos ฮธ = 0 & sin ฮธ = 1 Since, sin ฮธ is positive and cos ฮธ is zero Hence, ฮธ lies in Ist quadrant So, Argument = 90ยฐ = 90 ร— ๐œ‹/180 = ๐œ‹/2 Hence, argument of z = ๐œ‹/2 Example, 13 Find the modulus and argument of the complex numbers: (ii) 1/(1 + ๐‘–) First we simplify 1/(1 + ๐‘–) 1/(1 + ๐‘–) Rationalizing = 1/(1 + ๐‘–) ร— (1 โˆ’ ๐‘–)/(1 โˆ’ ๐‘–) = (1 ร— (1 โˆ’ ๐‘–))/(" " (1 + ๐‘–)(1 โˆ’ ๐‘–) ) Using (a โ€“ b) (a + b) = a2 โ€“ b2 = (1 โˆ’ ๐‘–)/((1)^2 โˆ’(๐‘–)^2 ) = (1 โˆ’๐‘–)/(1 โˆ’(โˆ’1) ) = (1 โˆ’ ๐‘–)/(1 + 1 ) = (1 โˆ’ ๐‘–)/2 = 1/2 + ๐‘– ((โˆ’ 1)/2) Now z = 1/2 + ๐‘– ((โˆ’ 1)/2) We calculate modulus by two different methods Method 1 To calculate Modulus of z z = 1/2 + ๐‘– ( (โˆ’ 1)/2 ) Complex number z is of the form ๐‘ฅ + ๐‘–๐‘ฆ Here ๐‘ฅ = 1/2 and ๐‘ฆ = (โˆ’ 1)/2 Modulus of z = |z| = โˆš(๐‘ฅ^2+๐‘ฆ2) = โˆš(( 1/2 )^2+( (โˆ’ 1)/2 )^2 ) = โˆš( 1/4+1/4 ) = โˆš( (1 + 1)/4 ) = โˆš( 2/4 ) = โˆš( 1/(" " 2" " )) = 1/(" " โˆš( 2) " " ) โ‡’ Modulus of ๐‘ง is 1/(" " โˆš( 2) " " ) Method 2 to calculate Modulus of z Given ๐‘ง = 1/2 + ๐‘– ( (โˆ’ 1)/2 ) Let z = ๐‘Ÿ(๐‘๐‘œ๐‘ โกฮธ+๐‘– sin ฮธ) Here r is modulus, and ฮธ is argument Form (1) and (2) 1/2 + ๐‘– ( (โˆ’ 1)/2 ) = ๐‘Ÿ(๐‘๐‘œ๐‘ โกฮธ+๐‘– sin ฮธ) 1/2 + ๐‘– ( (โˆ’ 1)/2 ) = rcos ฮธ + ๐‘– r sin ฮธ Adding (3) and (4) 1/4 + 1/4 = r2 cos2 ฮธ + r2 sin2 ฮธ (1 + 1)/4 = r2 ( cos2 ฮธ + sin2 ฮธ ) 2/4 = ๐‘Ÿ2 (cos2 ฮธ+sin2 ฮธ) 1/2 = ๐‘Ÿ2 ร— 1 โˆš(1/2) = r 1/โˆš2 = ๐‘Ÿ ๐‘Ÿ = 1/โˆš2 โ‡’ Modulus = 1/โˆš2 Finding argument 1/2 + ๐‘– ( (โˆ’ 1)/2 ) = rcos ฮธ + ๐‘– r sin ฮธ Comparing real part 1/2 = r cos ฮธ Put r = 1/โˆš2 1/2 = 1/โˆš2 cos ฮธ โˆš2/2 = cos ฮธ 1/โˆš2 = cos ฮธ โ‡’ cos ฮธ = 1/โˆš2 Hence, cos ฮธ = 1/โˆš2 & sin ฮธ = (โˆ’ 1)/โˆš2 Here, sin ฮธ is negative and cos ฮธ is positive, Hence, ฮธ lies in IVth quadrant So, Argument = โˆ’ 45ยฐ = โˆ’ 45ยฐ ร— ๐œ‹/(180ยฐ) = (โˆ’ ๐œ‹)/4 Hence, argument of ๐‘ง = (โˆ’ ๐œ‹)/4

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.