Example 13  - Chapter 5 Class 11 Complex Numbers - Part 8

Advertisement

Example 13  - Chapter 5 Class 11 Complex Numbers - Part 9

Advertisement

Example 13  - Chapter 5 Class 11 Complex Numbers - Part 10 Example 13  - Chapter 5 Class 11 Complex Numbers - Part 11 Example 13  - Chapter 5 Class 11 Complex Numbers - Part 12 Example 13  - Chapter 5 Class 11 Complex Numbers - Part 13 Example 13  - Chapter 5 Class 11 Complex Numbers - Part 14 Example 13  - Chapter 5 Class 11 Complex Numbers - Part 15

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

Transcript

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

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.