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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 + 𝑖)/(1 βˆ’ 𝑖) , First we solve (1 + 𝑖)/(1 βˆ’ 𝑖) Let 𝑧 = (1 + 𝑖)/(1 βˆ’ 𝑖) Rationalizing the same = (1 + 𝑖)/(1 βˆ’ 𝑖) Γ— (1 + 𝑖)/(1 + 𝑖) = (( 1 + 𝑖 ) ( 1 + 𝑖 ))/("(" 1 βˆ’ 𝑖 ) (1 + 𝑖 )) Using (a – b) (a + b) = a2 βˆ’ b2 = ( 1+ 𝑖 )2/( ( 1 )2 βˆ’ ( 𝑖 )2) Using ( a + b )2 = a2 + b2 + 2ab = ((1)2 + (𝑖)2 + 2𝑖)/( (1)2 βˆ’ (𝑖)2) Putting i2 = βˆ’ 1 = ((1)2 + (βˆ’1) + 2𝑖)/( 1βˆ’ (βˆ’ 1) ) = (1 βˆ’1 + 2𝑖)/( 1 + 1) = ( 2𝑖)/( 2) = 𝑖 = 0 + 𝑖 Hence, 𝑧 = 0 + 𝑖 Method 1 To calculate modulus of z z = 0 + i Complex number z is of the form x + 𝑖y Hence x = 0 and y = 1 Modulus of z = √(π‘₯^2+𝑦2) = √(( 0 )2+(1)2) = √(0+1) = √1 = 1 Modulus of z = 1 Method 2 To calculate modulus of z We have , z = 0 + 𝑖 Let z = r ( cos ΞΈ + 𝑖 sin ΞΈ ) Here r is modulus, and ΞΈ is argument From (1) and (2) 0 + 𝑖 = r ( cos ΞΈ + 𝑖 sin ΞΈ ) 0 + 𝑖 = r cos ΞΈ + 𝑖r sin ΞΈ Comparing real part 0 = r cos ΞΈ Squaring both sides (0)2 = ( π‘Ÿ cos⁑θ )2 0 = r2 cos2 ΞΈ Adding (3) and (4) 0 + 1 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 1=π‘Ÿ2 (cos2 ΞΈ+sin2 ΞΈ) 1 = r2 (1) 1 = r2 1 = r β‡’ Modulus of z = 1 Finding argument 0 + 𝑖 = r cos ΞΈ + 𝑖r sin ΞΈ Comparing real part 0 = r cos ΞΈ Put r = 1 0 = 1 Γ— cos ΞΈ 0 = cos ΞΈ cos ΞΈ = 0 Hence, cos ΞΈ = 0 & sin ΞΈ = 1 Since, sin ΞΈ is positive and cos ΞΈ is zero Hence, ΞΈ lies in Ist quadrant So, Argument = 90Β° = 90 Γ— πœ‹/180 = πœ‹/2 Hence, argument of z = πœ‹/2 Example, 13 Find the modulus and argument of the complex numbers: (ii) 1/(1 + 𝑖) First we simplify 1/(1 + 𝑖) 1/(1 + 𝑖) Rationalizing = 1/(1 + 𝑖) Γ— (1 βˆ’ 𝑖)/(1 βˆ’ 𝑖) = (1 Γ— (1 βˆ’ 𝑖))/(" " (1 + 𝑖)(1 βˆ’ 𝑖) ) Using (a – b) (a + b) = a2 – b2 = (1 βˆ’ 𝑖)/((1)^2 βˆ’(𝑖)^2 ) = (1 βˆ’π‘–)/(1 βˆ’(βˆ’1) ) = (1 βˆ’ 𝑖)/(1 + 1 ) = (1 βˆ’ 𝑖)/2 = 1/2 + 𝑖 ((βˆ’ 1)/2) Now z = 1/2 + 𝑖 ((βˆ’ 1)/2) We calculate modulus by two different methods Method 1 To calculate Modulus of z z = 1/2 + 𝑖 ( (βˆ’ 1)/2 ) Complex number z is of the form π‘₯ + 𝑖𝑦 Here π‘₯ = 1/2 and 𝑦 = (βˆ’ 1)/2 Modulus of z = |z| = √(π‘₯^2+𝑦2) = √(( 1/2 )^2+( (βˆ’ 1)/2 )^2 ) = √( 1/4+1/4 ) = √( (1 + 1)/4 ) = √( 2/4 ) = √( 1/(" " 2" " )) = 1/(" " √( 2) " " ) β‡’ Modulus of 𝑧 is 1/(" " √( 2) " " ) Method 2 to calculate Modulus of z Given 𝑧 = 1/2 + 𝑖 ( (βˆ’ 1)/2 ) Let z = π‘Ÿ(π‘π‘œπ‘ β‘ΞΈ+𝑖 sin ΞΈ) Here r is modulus, and ΞΈ is argument Form (1) and (2) 1/2 + 𝑖 ( (βˆ’ 1)/2 ) = π‘Ÿ(π‘π‘œπ‘ β‘ΞΈ+𝑖 sin ΞΈ) 1/2 + 𝑖 ( (βˆ’ 1)/2 ) = rcos ΞΈ + 𝑖 r sin ΞΈ Adding (3) and (4) 1/4 + 1/4 = r2 cos2 ΞΈ + r2 sin2 ΞΈ (1 + 1)/4 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 2/4 = π‘Ÿ2 (cos2 ΞΈ+sin2 ΞΈ) 1/2 = π‘Ÿ2 Γ— 1 √(1/2) = r 1/√2 = π‘Ÿ π‘Ÿ = 1/√2 β‡’ Modulus = 1/√2 Finding argument 1/2 + 𝑖 ( (βˆ’ 1)/2 ) = rcos ΞΈ + 𝑖 r sin ΞΈ Comparing real part 1/2 = r cos ΞΈ Put r = 1/√2 1/2 = 1/√2 cos ΞΈ √2/2 = cos ΞΈ 1/√2 = cos ΞΈ β‡’ cos ΞΈ = 1/√2 Hence, cos ΞΈ = 1/√2 & sin ΞΈ = (βˆ’ 1)/√2 Here, sin ΞΈ is negative and cos ΞΈ is positive, Hence, ΞΈ lies in IVth quadrant So, Argument = βˆ’ 45Β° = βˆ’ 45Β° Γ— πœ‹/(180Β°) = (βˆ’ πœ‹)/4 Hence, argument of 𝑧 = (βˆ’ πœ‹)/4

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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