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

 

 

 


Transcript

Example, 13 Find the modulus and argument of the complex numbers: (i) (1 + 𝑖)/(1 − 𝑖) , First we solve (1 + 𝑖)/(1 − 𝑖) Let 𝑧 = (1 + 𝑖)/(1 − 𝑖) Rationalizing the same = (1 + 𝑖)/(1 − 𝑖) × (1 + 𝑖)/(1 + 𝑖) = (( 1 + 𝑖 ) ( 1 + 𝑖 ))/("(" 1 − 𝑖 ) (1 + 𝑖 )) Using (a – b) (a + b) = a2 − b2 = ( 1+ 𝑖 )2/( ( 1 )2 − ( 𝑖 )2) Using ( a + b )2 = a2 + b2 + 2ab = ((1)2 + (𝑖)2 + 2𝑖)/( (1)2 − (𝑖)2) Putting i2 = − 1 = ((1)2 + (−1) + 2𝑖)/( 1− (− 1) ) = (1 −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 = √(𝑥^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 θ + 𝑖 sin θ ) Here r is modulus, and θ is argument From (1) and (2) 0 + 𝑖 = r ( cos θ + 𝑖 sin θ ) 0 + 𝑖 = r cos θ + 𝑖r sin θ Comparing real part 0 = r cos θ Squaring both sides (0)2 = ( 𝑟 cos⁡θ )2 0 = r2 cos2 θ Adding (3) and (4) 0 + 1 = r2 cos2 θ + r2 sin2 θ 1=𝑟2 (cos2 θ+sin2 θ) 1 = r2 (1) 1 = r2 1 = r ⇒ Modulus of z = 1 Finding argument 0 + 𝑖 = r cos θ + 𝑖r sin θ Comparing real part 0 = r cos θ Put r = 1 0 = 1 × cos θ 0 = cos θ cos θ = 0 Hence, cos θ = 0 & sin θ = 1 Since, sin θ is positive and cos θ is zero Hence, θ lies in Ist quadrant So, Argument = 90° = 90 × 𝜋/180 = 𝜋/2 Hence, argument of z = 𝜋/2

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.