Example 14 - Chapter 5 Class 11 Complex Numbers (Term 1)
Last updated at March 9, 2017 by Teachoo
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
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Example 14 Important You are here
Example 15
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Examples
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