Check sibling questions

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

Example 13  - Chapter 5 Class 11 Complex Numbers - Part 2
Example 13  - Chapter 5 Class 11 Complex Numbers - Part 3 Example 13  - Chapter 5 Class 11 Complex Numbers - Part 4 Example 13  - Chapter 5 Class 11 Complex Numbers - Part 5 Example 13  - Chapter 5 Class 11 Complex Numbers - Part 6 Example 13  - Chapter 5 Class 11 Complex Numbers - Part 7

This video is only available for Teachoo black users

Β 

Β 

Β 

Get Real time Doubt solving from 8pm to 12 am!


Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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.