# Example 14 - Chapter 5 Class 11 Complex Numbers (Term 1)

Last updated at March 9, 2017 by Teachoo

Examples

Example 1

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 2023 Exams

Example 8 Important Deleted for CBSE Board 2023 Exams

Example 9 Deleted for CBSE Board 2023 Exams

Example 10 Deleted for CBSE Board 2023 Exams

Example 11 Important Deleted for CBSE Board 2023 Exams

Example 12

Example 13 (i) Important

Example 13 (ii)

Example 14 Important You are here

Example 15

Example 16 Important Deleted for CBSE Board 2023 Exams

Last updated at March 9, 2017 by Teachoo

Example 14 If x + iy = (a + ib)(𝑎 − 𝑖𝑏) Taking R.H.S a + ib𝑎 − 𝑖𝑏 Rationalizing = a + ib𝑎 − 𝑖𝑏 (a + ib)(𝑎+ 𝑖𝑏) = (a + ib)2 𝑎 − 𝑖𝑏 (𝑎 + 𝑖𝑏) = 𝑎2 + (𝑖𝑏)2 + 2𝑎𝑖𝑏 𝑎2 − (𝑖𝑏)2 = 𝑎2 + 𝑖2 𝑏2 + 2𝑎𝑖𝑏 𝑎2 − 𝑖2 ( 𝑏2) = 𝑎2 − 𝑏2 + 2𝑎𝑖𝑏 𝑎2 + 𝑏2 = 𝑎2 − 𝑏2 𝑎2 + 𝑏2 + i 2𝑎𝑏 𝑎2 + 𝑏2 Now L.H.S = x + iy Comparing real and imaginary parts in L.H.S and R.H.S Adding (1) and (2) 𝑥2+ 𝑦2= 𝑎2 − 𝑏2 𝑎2 + 𝑏22+ (2𝑎𝑏)2 ( 𝑎2 + 𝑏2)𝟐 = 1 ( 𝑎2 + 𝑏2)𝟐 𝑎2 − 𝑏22 + 2𝑎𝑏2 = 1 ( 𝑎2 + 𝑏2)𝟐 𝑎4 + 𝑏4−2 𝑎2 𝑏2+ 4𝑎2 𝑏2 = 𝑎2 2 + 𝑏22 + 2 𝑎2 𝑏2 ( 𝑎2 + 𝑏2)𝟐 = ( 𝑎2 + 𝑏2)𝟐 ( 𝑎2 + 𝑏2)𝟐 = 1 Thus, 𝑥2+ 𝑦2=1 Hence, Proved