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

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

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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 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.