Check sibling questions

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

This video is only available for Teachoo black users

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

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

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.