Miscellaneous

Chapter 5 Class 11 Complex Numbers
Serial order wise

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Misc 4 If x β iy = β((a β ib)/(c β id)) prove that (π₯2 + π¦2)^2 = (a^2 + b^2)/(c^2 + d^2 ) Introduction (π₯ β ππ¦) (π₯+ ππ¦) Using ( a β b ) ( a + b ) = a2 β b2 = (π₯)^2 β (ππ¦)2 = π₯2 β (π) 2π¦2 = π₯2 β (β 1)π¦2 = π₯2 + π¦2 Misc 4 If x β iy = β((a β ib)/(c β id)) prove that (π₯2 + π¦2)^2 = (a^2 + b^2)/(c^2 + d^2 ) Given π₯ β ππ¦ = β((a β ib)/(c β id)) Calculating π₯ + ππ¦ Replacing β π by π π₯ + ππ¦ = β((a + ib)/( c + id)) Multiplying (1) &(2) (π₯ βππ¦) (π₯+ ππ¦) = β((a β ib)/(c β id)) Γ β((a + ib)/(c + id)) π₯2+π¦2 =β((aβib)/(cβid)Γ(a + ib)/(c + id)) =β((( a β ib) (a + ib))/((c β id) (c + id))) Using ( a β b ) ( a + b ) = a2 β b2 =β(((a)^2 β (ib)^2 )/((c)^2βγ (id)γ^2 )) =β((a^2 β i^2 b^2 )/(c^2 β i^2 d^2 )) Putting i2 = β1 =β((a2β(β1) b2 )/(c2β(β1)d2)) =β((a2+ b2 )/(c + d2)) Hence, π₯2 + π¦2 =β((a2+ b2 )/(c2 + d2)) Squaring both sides (x2 + y2)2 =(β((a2+ b2 )/(c2 + d2)))^2 (x2 + y2)2 = (a2+ b2 )/(c2 + d2) Hence Proved