Examples

Chapter 4 Class 11 Complex Numbers
Serial order wise

Β

Β

Β

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### Transcript

Example, 13 Find the modulus and argument of the complex numbers: (i) (1 + π)/(1 β π) , First we solve (1 + π)/(1 β π) Let π§ = (1 + π)/(1 β π) Rationalizing the same = (1 + π)/(1 β π) Γ (1 + π)/(1 + π) = (( 1 + π ) ( 1 + π ))/("(" 1 β π ) (1 + π )) Using (a β b) (a + b) = a2 β b2 = ( 1+ π )2/( ( 1 )2 β ( π )2) Using ( a + b )2 = a2 + b2 + 2ab = ((1)2 + (π)2 + 2π)/( (1)2 β (π)2) Putting i2 = β 1 = ((1)2 + (β1) + 2π)/( 1β (β 1) ) = (1 β1 + 2π)/( 1 + 1) = ( 2π)/( 2) = π = 0 + π Hence, π§ = 0 + π Method 1 To calculate modulus of z z = 0 + i Complex number z is of the form x + πy Hence x = 0 and y = 1 Modulus of z = β(π₯^2+π¦2) = β(( 0 )2+(1)2) = β(0+1) = β1 = 1 Modulus of z = 1 Method 2 To calculate modulus of z We have , z = 0 + π Let z = r ( cos ΞΈ + π sin ΞΈ ) Here r is modulus, and ΞΈ is argument From (1) and (2) 0 + π = r ( cos ΞΈ + π sin ΞΈ ) 0 + π = r cos ΞΈ + πr sin ΞΈ Comparing real part 0 = r cos ΞΈ Squaring both sides (0)2 = ( π cosβ‘ΞΈ )2 0 = r2 cos2 ΞΈ Adding (3) and (4) 0 + 1 = r2 cos2 ΞΈ + r2 sin2 ΞΈ 1=π2 (cos2 ΞΈ+sin2 ΞΈ) 1 = r2 (1) 1 = r2 1 = r β Modulus of z = 1 Finding argument 0 + π = r cos ΞΈ + πr 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/180 = π/2 Hence, argument of z = π/2