Proof- Replacing iota with -iota

Chapter 5 Class 11 Complex Numbers
Concept wise

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Misc 11 (Method 1) If a + ib = (x + π)2/(2x^2 + 1) , prove that π2 + π2 = (x^2+ 1)2/(2x^2+ 1)^2 π + ππ = (x + i)2/(2x2+ 1) Using ( π + π )^2 = π2 + π2 + 2ππ = (π₯2 + (π)^2 + 2π₯π)/(2π₯2+1) Putting π2 = β1 = (π₯2 β 1 + 2π₯π)/(2π₯2+ 1) = (x2 β 1)/(2x2 + 1) + π 2x/(2x2 + 1) Hence π + ππ = (x2 β 1)/(2x2 + 1) + π 2x/(2x2 + 1) Comparing real part π = (π₯^2 β 1)/(2π₯^2 + 1) Comparing imaginary part b = 2π₯/(2π₯2 + 1) Calculating π2 + π2 π2 + π2 = ((π₯^2 β 1)/(2π₯2 + 1))^2 + (2π₯/(2π₯2 + 1))^2 = ((π₯2β 1)2 + (2π₯)2)/((2π₯2 + 1)2) Using (π β π)^2 = π2 + π2 β 2ππ = ((π₯2 )2 + (1)2 β 2( π₯2)1 + 4π₯2)/( (2π₯2 + 1)2) = (π₯4 + 1 β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 + π )^2 = π2 + π2 + 2ππ = (π₯2+ 1)2/((2π₯2 + 1)2) Hence π2 + π2 = (π₯2+ 1)2/((2π₯2 + 1)2) Hence proved Misc 11 (Method 2) If a + ib = (x + π)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Introduction (π + ππ) ( π β ππ) Using ( a β b ) ( a + b ) = a2 β b2 = π2 β (ππ)2 = π2 β π2π2 Putting i2 = β1 = π2β (β1) π2 = π2 + π2 Hence, (π + ππ) (π β ππ) = π2 + π2 Misc 11 (Method 2) If a + ib = (x + π)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Given π + ππ = (π₯ + π)2/(2π₯2 + 1) For π β ππ Replace π by β π in (1) π β ππ = (π₯ β π)2/(2π₯2 + 1) Calculating (π β ππ) (π + ππ) (π β ππ) (π + ππ) = (π₯ β π)2/(2π₯2 + 1) Γ (π₯ + π)2/(2π₯2 + 1) π2 + π2 = ((π₯ β π)2 (π₯ + π)2)/(2π₯2 +1)2 = ( (π₯ β π) (π₯ + π))^2/(2π₯2 +1)2 Using ( a β b ) ( a + b ) = a2 β b2 = (( π₯^2 β (π)^2 )^2 )/(2π₯^2 + 1)2 = γ( π₯2β (β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