1. Chapter 5 Class 11 Complex Numbers
2. Serial order wise

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Misc 11 (Method 1) If a + ib = (x + ๐)2/(2x^2 + 1) , prove that ๐2 + ๐2 = (x^2+ 1)2/(2x^2+ 1)^2 ๐ + ๐๐ = (x + i)2/(2x2+ 1) Using ( ๐ + ๐ )^2 = ๐2 + ๐2 + 2๐๐ = (๐ฅ2 + (๐)^2 + 2๐ฅ๐)/(2๐ฅ2+1) Putting ๐2 = โ1 = (๐ฅ2 โ 1 + 2๐ฅ๐)/(2๐ฅ2+ 1) = (x2 โ 1)/(2x2 + 1) + ๐ 2x/(2x2 + 1) Hence ๐ + ๐๐ = (x2 โ 1)/(2x2 + 1) + ๐ 2x/(2x2 + 1) Comparing real part ๐ = (๐ฅ^2 โ 1)/(2๐ฅ^2 + 1) Comparing imaginary part b = 2๐ฅ/(2๐ฅ2 + 1) Calculating ๐2 + ๐2 ๐2 + ๐2 = ((๐ฅ^2 โ 1)/(2๐ฅ2 + 1))^2 + (2๐ฅ/(2๐ฅ2 + 1))^2 = ((๐ฅ2โ 1)2 + (2๐ฅ)2)/((2๐ฅ2 + 1)2) Using (๐ โ ๐)^2 = ๐2 + ๐2 โ 2๐๐ = ((๐ฅ2 )2 + (1)2 โ 2( ๐ฅ2)1 + 4๐ฅ2)/( (2๐ฅ2 + 1)2) = (๐ฅ4 + 1 โ2๐ฅ2 + 4๐ฅ2)/((2๐ฅ2 +1)2) = (๐ฅ4 + 1 + 2๐ฅ2)/((2๐ฅ2 + 1)2) = ((๐ฅ2)2 + (1)2 + 2(๐ฅ2) (1))/((2๐ฅ^2 + 1)2) Using ( ๐ + ๐ )^2 = ๐2 + ๐2 + 2๐๐ = (๐ฅ2+ 1)2/((2๐ฅ2 + 1)2) Hence ๐2 + ๐2 = (๐ฅ2+ 1)2/((2๐ฅ2 + 1)2) Hence proved Misc 11 (Method 2) If a + ib = (x + ๐)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Introduction (๐ + ๐๐) ( ๐ โ ๐๐) Using ( a โ b ) ( a + b ) = a2 โ b2 = ๐2 โ (๐๐)2 = ๐2 โ ๐2๐2 Putting i2 = โ1 = ๐2โ (โ1) ๐2 = ๐2 + ๐2 Hence, (๐ + ๐๐) (๐ โ ๐๐) = ๐2 + ๐2 Misc 11 (Method 2) If a + ib = (x + ๐)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Given ๐ + ๐๐ = (๐ฅ + ๐)2/(2๐ฅ2 + 1) For ๐ โ ๐๐ Replace ๐ by โ ๐ in (1) ๐ โ ๐๐ = (๐ฅ โ ๐)2/(2๐ฅ2 + 1) Calculating (๐ โ ๐๐) (๐ + ๐๐) (๐ โ ๐๐) (๐ + ๐๐) = (๐ฅ โ ๐)2/(2๐ฅ2 + 1) ร (๐ฅ + ๐)2/(2๐ฅ2 + 1) ๐2 + ๐2 = ((๐ฅ โ ๐)2 (๐ฅ + ๐)2)/(2๐ฅ2 +1)2 = ( (๐ฅ โ ๐) (๐ฅ + ๐))^2/(2๐ฅ2 +1)2 Using ( a โ b ) ( a + b ) = a2 โ b2 = (( ๐ฅ^2 โ (๐)^2 )^2 )/(2๐ฅ^2 + 1)2 = ใ( ๐ฅ2โ (โ1)) ใ^2/(2๐ฅ2 + 1)2 = ( ๐ฅ2 + 1)2/(2๐ฅ2 + 1)2 Hence a2 + b2 = (๐ฅ2 + 1 )/(2๐ฅ2 + 1) Hence proved