Misc 19 - If (a + ib) (c + id) (e + if) (g + ih) = A + iB - Miscellaneous

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Misc 19 If (๐‘Ž+๐‘–๐‘)(๐‘+๐‘–๐‘‘)(๐‘’+๐‘–๐‘“)(๐‘”+๐‘–โ„Ž)=๐ด+๐‘–๐ต, then show that (๐‘Ž2 + ๐‘2) (๐‘2 + ๐‘‘2) (๐‘’2 + ๐‘“2) (๐‘”2 + โ„Ž2) = ๐ด2 +๐ต2. Introduction (๐ด + ๐‘–๐ต) ( ๐ด โ€“ ๐‘–๐ต) Using ( a โ€“ b ) ( a + b ) = a2 โ€“ b2 = ๐ด2 โ€“ (๐‘–๐ต)2 = ๐ด2 โ€“ ๐‘–2 ๐ต2 Putting i2 = โˆ’1 = ๐ด2 โ€“ ( โˆ’1) ๐ต2 = ๐ด2 +๐ต2 Hence, (๐ด + ๐‘–๐ต) (๐ด โ€“ ๐‘–๐ต) = ๐ด2 +๐ต2 Misc, 19 If (๐‘Ž+๐‘–๐‘)(๐‘+๐‘–๐‘‘)(๐‘’+๐‘–๐‘“)(๐‘”+๐‘–โ„Ž)=๐ด+๐‘–๐ต, then show that (๐‘Ž2 + ๐‘2) (๐‘2 + ๐‘‘2) (๐‘’2 + ๐‘“2) (๐‘”2 + โ„Ž2) = ๐ด2 +๐ต2. Given ( ๐ด + ๐‘–๐ต ) = (๐‘Ž + ๐‘–๐‘ ) ( ๐‘ + ๐‘–๐‘‘ ) (๐‘’ + ๐‘–๐‘“ ) ( ๐‘” + ๐‘–โ„Ž ) To calculate ( ๐ด โ€“ ๐‘–๐ต ) Replacing ๐‘– by โ€“๐‘– in (1) (๐ด โˆ’๐‘–๐ต ) = ( ๐‘Ž โ€“ ๐‘–๐‘ ) ( ๐‘ โ€“ ๐‘–๐‘‘ ) ( ๐‘’ โ€“ ๐‘–๐‘“ ) ( ๐‘” โ€“ ๐‘–โ„Ž ) Now, calculating (๐ด + ๐‘–๐ต) ( ๐ด โ€“ ๐‘–๐ต) (๐ด + ๐‘–๐ต) ( ๐ด โ€“ ๐‘–๐ต) = (๐‘Ž + ๐‘–๐‘ )( ๐‘ + ๐‘–๐‘‘ )(๐‘’ + ๐‘–๐‘“ )( ๐‘” + ๐‘–โ„Ž )(๐‘Ž โˆ’ ๐‘–๐‘ ) ( ๐‘ โˆ’ ๐‘–๐‘‘ ) (๐‘’ โˆ’ ๐‘–๐‘“ ) ( ๐‘” โˆ’ ๐‘–โ„Ž ) ๐ด2 + ๐ต^2= [( ๐‘Ž+ ๐‘–๐‘ )(๐‘Ž โ€“ ๐‘–๐‘ )][(๐‘+ ๐‘–๐‘‘)(๐‘ โ€“ ๐‘–๐‘‘ )] [( ๐‘’ + ๐‘–๐‘“) ( ๐‘’ โ€“ ๐‘–๐‘“ )] [( ๐‘” + ๐‘–โ„Ž ) ( ๐‘” โ€“ ๐‘–โ„Ž)] ๐‘ˆ๐‘ ๐‘–๐‘›๐‘” ( ๐‘ฅ โ€“ ๐‘ฆ ) ( ๐‘ฅ + ๐‘ฆ ) = ๐‘ฅ2+๐‘ฆ2 = [(๐‘Ž)^2 โ€“ (๐‘–๐‘)2] [ ๐‘2 โ€“ ( ๐‘–๐‘‘)^2] [๐‘’2โˆ’ (๐‘–๐‘“)^2 ] [๐‘”2 โ€“ (โˆ’ ๐‘–โ„Ž)]2 = [ ๐‘Ž2 โˆ’ ๐‘2 ๐‘–2 ] [ ๐‘2 โˆ’ ๐‘–2 ๐‘‘2 ] [ ๐‘’2 โˆ’ ๐‘–2 ๐‘“2 ] [ ๐‘”2 โˆ’ ๐‘–2 โ„Ž2 ] Putting i2 = โˆ’1 = [ ๐‘Ž2โ€“ (โˆ’1)๐‘2] [ ๐‘2 โ€“ (โˆ’1) ๐‘‘ ] [ ๐‘’2 โ€“ (โˆ’1) ๐‘“)] [๐‘”2 โ€“ (โˆ’1) โ„Ž2 ] = [ ๐‘Ž2 + ๐‘2 ] [ ๐‘2 + ๐‘‘2 ] [ ๐‘’2 + ๐‘“2 ] [๐‘”2 + โ„Ž2 ] Hence, (๐‘Ž2 + ๐‘2) (๐‘2 + ๐‘‘2) (๐‘’2 + ๐‘“2) (๐‘”2 + โ„Ž2) = ๐ด2 +๐ต2. Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.