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Question 1 (i) Deleted for CBSE Board 2025 Exams

Question 1 (ii) Important Deleted for CBSE Board 2025 Exams

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Question 6 Important Deleted for CBSE Board 2025 Exams You are here

Chapter 4 Class 11 Complex Numbers

Serial order wise

Last updated at April 16, 2024 by Teachoo

Question 6 Find the modulus and argument of the complex number ( 1 + 2i)/(1 β 3i) . First we solve ( 1 + 2π)/(1 β 3π) Let π§ = ( 1 + 2π)/(1 β 3π) Rationalizing the same = ( 1 + 2π)/(1 β 3π) x ( 1 + 3π)/(1 + 3π) = ((1 + 2π) ( 1 + 3π))/((1 β 3π) ( 1 + 3π)) = (1 (1 + 3π) + 2π (1 + 3π))/((1 β 3π) (1 + 3π) ) = ( 1 + 3π+ 2π+ 6π2)/((1 β 3π) (1 + 3π) ) Using ( a β b ) ( a + b ) = a2 - b2 = ( 1+ 5π+ 6π2)/((1)2 β (3π)2) = (1 + 5π+ 6π2)/(1 β 9π2) Putting π2 = - 1 = (1 + 5π + 6 (β1))/(1 β 9 (β1)) = (1 + 5π β 6 )/(1 + 9 ) = ( \β 5 + 5π)/10 = ( 5 (β1 +π))/10 = (β 1+ π)/2 = (β 1)/2+π ( 1/2 ) Thus, π§ = (β 1)/2+π ( 1/2 ) Method 1 To find Modulus Now we have z = (β 1)/2 + π (1/2) Complex number z is of the form x + π y Here x = (β 1)/2 and y = 1/2 Modulus of z = |z| = β(π₯^2+π¦2) = β(( (β 1)/(2 ))^2+( 1/(2 ))^2 ) = β(1/4+1/4) = β(2/4) = β(1/2) = 1/β2 Modulus of z = 1/β2 Method 2 To calculate modulus of z Given z = (β1)/2 + π (1/2) Let π§=π (cosβ‘ΞΈ+π sinΞΈ) Here r is modulus, and ΞΈ is argument Form (1) and (2) (β1)/2 + π (1/2) = π (cosβ‘ΞΈ+π sinΞΈ) (β1)/2 + π (1/2) = r cos ΞΈ + πr sin ΞΈ Comparing real part (β 1)/2 = r cos 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 = r2 Γ 1 1/2 = r2 1/β2 = r Modulus of z = 1/β2 Finding argument (β1)/2 + π (1/2) = r cos ΞΈ + πr sin ΞΈ Hence, sin ΞΈ = 1/β2 & cos ΞΈ = (β 1)/β2 Here, sin ΞΈ is positive and cos ΞΈ is negative, Hence, ΞΈ lies in IVth quadrant Argument = 180Β° β 45Β° = 135Β° = 135Β° Γ π/180o = ( 3 π)/4 So argument of z = ( 3 π)/4