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Misc 11 - If a + ib = (x + i)2/(2x2 + 1), prove a2 + b2 - Miscellaneous

Misc 11  - Chapter 5 Class 11 Complex Numbers - Part 2
Misc 11  - Chapter 5 Class 11 Complex Numbers - Part 3

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Misc 11  - Chapter 5 Class 11 Complex Numbers - Part 4
Misc 11  - Chapter 5 Class 11 Complex Numbers - Part 5
Misc 11  - Chapter 5 Class 11 Complex Numbers - Part 6

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Misc 11 (Method 1) If a + ib = (x + 𝑖)2/(2x^2 + 1) , prove that π‘Ž2 + 𝑏2 = (x^2+ 1)2/(2x^2+ 1)^2 π‘Ž + 𝑖𝑏 = (x + i)2/(2x2+ 1) Using ( π‘Ž + 𝑏 )^2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘ = (π‘₯2 + (𝑖)^2 + 2π‘₯𝑖)/(2π‘₯2+1) Putting 𝑖2 = βˆ’1 = (π‘₯2 βˆ’ 1 + 2π‘₯𝑖)/(2π‘₯2+ 1) = (x2 βˆ’ 1)/(2x2 + 1) + 𝑖 2x/(2x2 + 1) Hence π‘Ž + 𝑖𝑏 = (x2 βˆ’ 1)/(2x2 + 1) + 𝑖 2x/(2x2 + 1) Comparing real part π‘Ž = (π‘₯^2 βˆ’ 1)/(2π‘₯^2 + 1) Comparing imaginary part b = 2π‘₯/(2π‘₯2 + 1) Calculating π‘Ž2 + 𝑏2 π‘Ž2 + 𝑏2 = ((π‘₯^2 βˆ’ 1)/(2π‘₯2 + 1))^2 + (2π‘₯/(2π‘₯2 + 1))^2 = ((π‘₯2βˆ’ 1)2 + (2π‘₯)2)/((2π‘₯2 + 1)2) Using (π‘Ž βˆ’ 𝑏)^2 = π‘Ž2 + 𝑏2 βˆ’ 2π‘Žπ‘ = ((π‘₯2 )2 + (1)2 βˆ’ 2( π‘₯2)1 + 4π‘₯2)/( (2π‘₯2 + 1)2) = (π‘₯4 + 1 βˆ’2π‘₯2 + 4π‘₯2)/((2π‘₯2 +1)2) = (π‘₯4 + 1 + 2π‘₯2)/((2π‘₯2 + 1)2) = ((π‘₯2)2 + (1)2 + 2(π‘₯2) (1))/((2π‘₯^2 + 1)2) Using ( π‘Ž + 𝑏 )^2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘ = (π‘₯2+ 1)2/((2π‘₯2 + 1)2) Hence π‘Ž2 + 𝑏2 = (π‘₯2+ 1)2/((2π‘₯2 + 1)2) Hence proved Misc 11 (Method 2) If a + ib = (x + 𝑖)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Introduction (π‘Ž + 𝑖𝑏) ( π‘Ž – 𝑖𝑏) Using ( a – b ) ( a + b ) = a2 – b2 = π‘Ž2 – (𝑖𝑏)2 = π‘Ž2 – 𝑖2𝑏2 Putting i2 = βˆ’1 = π‘Ž2βˆ’ (βˆ’1) 𝑏2 = π‘Ž2 + 𝑏2 Hence, (π‘Ž + 𝑖𝑏) (π‘Ž – 𝑖𝑏) = π‘Ž2 + 𝑏2 Misc 11 (Method 2) If a + ib = (x + 𝑖)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Given π‘Ž + 𝑖𝑏 = (π‘₯ + 𝑖)2/(2π‘₯2 + 1) For π‘Ž – 𝑖𝑏 Replace 𝑖 by – 𝑖 in (1) π‘Ž – 𝑖𝑏 = (π‘₯ βˆ’ 𝑖)2/(2π‘₯2 + 1) Calculating (π‘Ž – 𝑖𝑏) (π‘Ž + 𝑖𝑏) (π‘Ž – 𝑖𝑏) (π‘Ž + 𝑖𝑏) = (π‘₯ βˆ’ 𝑖)2/(2π‘₯2 + 1) Γ— (π‘₯ + 𝑖)2/(2π‘₯2 + 1) π‘Ž2 + 𝑏2 = ((π‘₯ βˆ’ 𝑖)2 (π‘₯ + 𝑖)2)/(2π‘₯2 +1)2 = ( (π‘₯ βˆ’ 𝑖) (π‘₯ + 𝑖))^2/(2π‘₯2 +1)2 Using ( a – b ) ( a + b ) = a2 – b2 = (( π‘₯^2 βˆ’ (𝑖)^2 )^2 )/(2π‘₯^2 + 1)2 = γ€–( π‘₯2βˆ’ (βˆ’1)) γ€—^2/(2π‘₯2 + 1)2 = ( π‘₯2 + 1)2/(2π‘₯2 + 1)2 Hence a2 + b2 = (π‘₯2 + 1 )/(2π‘₯2 + 1) Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.