Miscellaneous

Misc 1
Important

Misc 2

Misc 3

Misc 4 Important

Misc 5 (i) Deleted for CBSE Board 2022 Exams

Misc 5 (ii) Important

Misc 6

Misc 7

Misc 8 Important

Misc 9

Misc 10 Important You are here

Misc 11

Misc 12

Misc 13 Important Deleted for CBSE Board 2022 Exams

Misc 14

Misc 15 Important

Misc 16

Misc 17 Important

Misc 18 Important

Misc 19

Misc 20 Important

Chapter 5 Class 11 Complex Numbers (Term 1)

Serial order wise

Last updated at Jan. 10, 2019 by Teachoo

Misc 10 (Method 1) If z1 = 2 β i, z2 = 1 + i, find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| We have to find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| First we find (π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1) (π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1) = ("(" 2"β" π")" + "(" 1+ π") " + 1)/("(" 2"β" π")" β" (" 1+ π")" + 1) = (2 β π + 1 + π + 1)/(2 β π β 1 β" " π + 1) = (2 + 1 + 1 β π + π )/(2 β 1 + 1 β π β" " π) = (4 + 0)/(2 β2π ) = 4/(2 (1 β π) ) = 4/(2 (1 β π) ) = 2/((1 β π) ) Rationalizing = 2/(1 β π) Γ (1 + π)/(1 + π) = (2 (1 + π))/((1 β π) (1 + π)) Using (a β b) (a + b) = a2 - b2 = (2(1 + π))/((1)2 β (π)2) Putting i2 = β1 = (2 (1 + π ))/(1 β(β1) ) = (2(1 + π))/(1 + 1) = (2 (1 + π))/2 = 1 + π Hence, (π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1) = 1 + π Now we find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| i.e. |1 + π| Complex number z is of the form π₯ + π π¦ Here x = 1 and y = 1 Modulus of z = |z| = β(π₯^2+π¦2) = β((1)2+( 1)2) = β(1+1) = β2 So, |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)|= β2 Misc 10 (Method 2) If z1 = 2 β i, z2 = 1 + i, find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| We have , z = 1 + π Let Polar form of z = r ( cos ΞΈ + π sin ΞΈ ) From (1) and (2) 1 + π (1) = π (cosβ‘ΞΈ + π sin ΞΈ ) 1 + π (1) = π cosβ‘ΞΈ + πr sin ΞΈ Comparing real part 1 = r cos ΞΈ Squaring both side (1)2 = (π cosΞΈ) 1 = r2 cos2 ΞΈ r2 cos2 ΞΈ = 1 Comparing Imaginary parts 1 = rγ sinγβ‘ΞΈ Squaring both sides (1)2 = ( r2 sin ΞΈ )2 1 = r2 sin2β‘ΞΈ r2 sin2β‘ΞΈ = 1 Adding (3) and (4) 1 + 1 = π2 cos2 ΞΈ + π2 sin2 ΞΈ 2 = π2 (cos2 ΞΈ + sin2 ΞΈ) 2 = r2 Γ 1 2 = r2 β2 = r r = β2 Modulus of π§ = β2