Examples

Chapter 4 Class 11 Complex Numbers
Serial order wise

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Example, 13 Find the modulus and argument of the complex numbers: (ii) 1/(1 + π) First we simplify 1/(1 + π) 1/(1 + π) Rationalizing = 1/(1 + π) Γ (1 β π)/(1 β π) = (1 Γ (1 β π))/(" " (1 + π)(1 β π) ) Using (a β b) (a + b) = a2 β b2 = (1 β π)/((1)^2 β(π)^2 ) = (1 βπ)/(1 β(β1) ) = (1 β π)/(1 + 1 ) = (1 β π)/2 = 1/2 + π ((β 1)/2) Now z = 1/2 + π ((β 1)/2) We calculate modulus by two different methods Method 1 To calculate Modulus of z z = 1/2 + π ( (β 1)/2 ) Complex number z is of the form π₯ + ππ¦ Here π₯ = 1/2 and π¦ = (β 1)/2 Modulus of z = |z| = β(π₯^2+π¦2) = β(( 1/2 )^2+( (β 1)/2 )^2 ) = β( 1/4+1/4 ) = β( (1 + 1)/4 ) = β( 2/4 ) = β( 1/(" " 2" " )) = 1/(" " β( 2) " " ) β Modulus of π§ is 1/(" " β( 2) " " ) Method 2 to calculate Modulus of z Given π§ = 1/2 + π ( (β 1)/2 ) Let z = π(πππ β‘ΞΈ+π sin ΞΈ) Here r is modulus, and ΞΈ is argument Form (1) and (2) 1/2 + π ( (β 1)/2 ) = π(πππ β‘ΞΈ+π sin ΞΈ) 1/2 + π ( (β 1)/2 ) = rcos ΞΈ + π r sin ΞΈ Adding (3) and (4) 1/4 + 1/4 = r2 cos2 ΞΈ + r2 sin2 ΞΈ (1 + 1)/4 = r2 ( cos2 ΞΈ + sin2 ΞΈ ) 2/4 = π2 (cos2 ΞΈ+sin2 ΞΈ) 1/2 = π2 Γ 1 β(1/2) = r 1/β2 = π π = 1/β2 β Modulus = 1/β2 Finding argument 1/2 + π ( (β 1)/2 ) = rcos ΞΈ + π r sin ΞΈ Comparing real part 1/2 = r cos ΞΈ Put r = 1/β2 1/2 = 1/β2 cos ΞΈ β2/2 = cos ΞΈ 1/β2 = cos ΞΈ β cos ΞΈ = 1/β2 Hence, cos ΞΈ = 1/β2 & sin ΞΈ = (β 1)/β2 Here, sin ΞΈ is negative and cos ΞΈ is positive, Hence, ΞΈ lies in IVth quadrant So, Argument = β 45Β° = β 45Β° Γ π/(180Β°) = (β π)/4 Hence, argument of π§ = (β π)/4