Β
Β
Β
Examples
Example 2 (i)
Example 2 (ii) Important
Example 3
Example 4
Example 5 Important
Example 6 (i)
Example 6 (ii) Important
Example 7 Deleted for CBSE Board 2022 Exams
Example 8 Important Deleted for CBSE Board 2022 Exams
Example 9
Example 10
Example 11 Important
Example 12
Example 13 (i) Important Deleted for CBSE Board 2022 Exams You are here
Example 13 (ii) Deleted for CBSE Board 2022 Exams
Example 14 Important
Example 15
Example 16 Important Deleted for CBSE Board 2022 Exams
Examples
Last updated at Sept. 3, 2021 by Teachoo
Β
Β
Β
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 + (β1) + 2π)/( 1β (β 1) ) = (1 β1 + 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 + πy Hence x = 0 and y = 1 Modulus of z = β(π₯^2+π¦2) = β(( 0 )2+(1)2) = β(0+1) = β1 = 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 + πr 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=π2 (cos2 ΞΈ+sin2 ΞΈ) 1 = r2 (1) 1 = r2 1 = r β Modulus of z = 1 Finding argument 0 + π = r cos ΞΈ + πr 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