Example 13 (i) - Chapter 5 Class 11 Complex Numbers (Term 1)

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