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Misc 6(Method 1) If a + ib = (x + 𝑖)2/(2x^2 + 1) , prove that 𝑎2 + 𝑏2 = (x^2+ 1)2/(2x^2+ 1)^2 𝑎 + 𝑖𝑏 = (x + i)2/(2x2+ 1) Using ( 𝑎 + 𝑏 )^2 = 𝑎2 + 𝑏2 + 2𝑎𝑏 = (𝑥2 + (𝑖)^2 + 2𝑥𝑖)/(2𝑥2+1) Putting 𝑖2 = −1 = (𝑥2 − 1 + 2𝑥𝑖)/(2𝑥2+ 1) = (x2 − 1)/(2x2 + 1) + 𝑖 2x/(2x2 + 1) Hence 𝑎 + 𝑖𝑏 = (x2 − 1)/(2x2 + 1) + 𝑖 2x/(2x2 + 1) Comparing real part 𝑎 = (𝑥^2 − 1)/(2𝑥^2 + 1) Comparing imaginary part b = 2𝑥/(2𝑥2 + 1) Calculating 𝑎2 + 𝑏2 𝑎2 + 𝑏2 = ((𝑥^2 − 1)/(2𝑥2 + 1))^2 + (2𝑥/(2𝑥2 + 1))^2 = ((𝑥2− 1)2 + (2𝑥)2)/((2𝑥2 + 1)2) Using (𝑎 − 𝑏)^2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 = ((𝑥2 )2 + (1)2 − 2( 𝑥2)1 + 4𝑥2)/( (2𝑥2 + 1)2) = (𝑥4 + 1 −2𝑥2 + 4𝑥2)/((2𝑥2 +1)2) = (𝑥4 + 1 + 2𝑥2)/((2𝑥2 + 1)2) = ((𝑥2)2 + (1)2 + 2(𝑥2) (1))/((2𝑥^2 + 1)2) Using ( 𝑎 + 𝑏 )^2 = 𝑎2 + 𝑏2 + 2𝑎𝑏 = (𝑥2+ 1)2/((2𝑥2 + 1)2) Hence 𝑎2 + 𝑏2 = (𝑥2+ 1)2/((2𝑥2 + 1)2) Hence proved Misc 6(Method 2) If a + ib = (x + 𝑖)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Introduction (𝑎 + 𝑖𝑏) ( 𝑎 – 𝑖𝑏) Using ( a – b ) ( a + b ) = a2 – b2 = 𝑎2 – (𝑖𝑏)2 = 𝑎2 – 𝑖2𝑏2 Putting i2 = −1 = 𝑎2− (−1) 𝑏2 = 𝑎2 + 𝑏2 Hence, (𝑎 + 𝑖𝑏) (𝑎 – 𝑖𝑏) = 𝑎2 + 𝑏2 Misc 6(Method 2) If a + ib = (x + 𝑖)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Given 𝑎 + 𝑖𝑏 = (𝑥 + 𝑖)2/(2𝑥2 + 1) For 𝑎 – 𝑖𝑏 Replace 𝑖 by – 𝑖 in (1) 𝑎 – 𝑖𝑏 = (𝑥 − 𝑖)2/(2𝑥2 + 1) Calculating (𝑎 – 𝑖𝑏) (𝑎 + 𝑖𝑏) (𝑎 – 𝑖𝑏) (𝑎 + 𝑖𝑏) = (𝑥 − 𝑖)2/(2𝑥2 + 1) × (𝑥 + 𝑖)2/(2𝑥2 + 1) 𝑎2 + 𝑏2 = ((𝑥 − 𝑖)2 (𝑥 + 𝑖)2)/(2𝑥2 +1)2 = ( (𝑥 − 𝑖) (𝑥 + 𝑖))^2/(2𝑥2 +1)2 Using ( a – b ) ( a + b ) = a2 – b2 = (( 𝑥^2 − (𝑖)^2 )^2 )/(2𝑥^2 + 1)2 = 〖( 𝑥2− (−1)) 〗^2/(2𝑥2 + 1)2 = ( 𝑥2 + 1)2/(2𝑥2 + 1)2 Hence a2 + b2 = (𝑥2 + 1 )/(2𝑥2 + 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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.