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

Misc 19 - Chapter 5 Class 11 Complex Numbers - Part 2
Misc 19 - Chapter 5 Class 11 Complex Numbers - Part 3


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Misc 19 If (π‘Ž+𝑖𝑏)(𝑐+𝑖𝑑)(𝑒+𝑖𝑓)(𝑔+π‘–β„Ž)=𝐴+𝑖𝐡, then show that (π‘Ž2 + 𝑏2) (𝑐2 + 𝑑2) (𝑒2 + 𝑓2) (𝑔2 + β„Ž2) = 𝐴2 +𝐡2. Introduction (𝐴 + 𝑖𝐡) ( 𝐴 – 𝑖𝐡) Using ( a – b ) ( a + b ) = a2 – b2 = 𝐴2 – (𝑖𝐡)2 = 𝐴2 – 𝑖2 𝐡2 Putting i2 = βˆ’1 = 𝐴2 – ( βˆ’1) 𝐡2 = 𝐴2 +𝐡2 Hence, (𝐴 + 𝑖𝐡) (𝐴 – 𝑖𝐡) = 𝐴2 +𝐡2 Misc, 19 If (π‘Ž+𝑖𝑏)(𝑐+𝑖𝑑)(𝑒+𝑖𝑓)(𝑔+π‘–β„Ž)=𝐴+𝑖𝐡, then show that (π‘Ž2 + 𝑏2) (𝑐2 + 𝑑2) (𝑒2 + 𝑓2) (𝑔2 + β„Ž2) = 𝐴2 +𝐡2. Given ( 𝐴 + 𝑖𝐡 ) = (π‘Ž + 𝑖𝑏 ) ( 𝑐 + 𝑖𝑑 ) (𝑒 + 𝑖𝑓 ) ( 𝑔 + π‘–β„Ž ) To calculate ( 𝐴 – 𝑖𝐡 ) Replacing 𝑖 by –𝑖 in (1) (𝐴 βˆ’π‘–π΅ ) = ( π‘Ž – 𝑖𝑏 ) ( 𝑐 – 𝑖𝑑 ) ( 𝑒 – 𝑖𝑓 ) ( 𝑔 – π‘–β„Ž ) Now, calculating (𝐴 + 𝑖𝐡) ( 𝐴 – 𝑖𝐡) (𝐴 + 𝑖𝐡) ( 𝐴 – 𝑖𝐡) = (π‘Ž + 𝑖𝑏 )( 𝑐 + 𝑖𝑑 )(𝑒 + 𝑖𝑓 )( 𝑔 + π‘–β„Ž )(π‘Ž βˆ’ 𝑖𝑏 ) ( 𝑐 βˆ’ 𝑖𝑑 ) (𝑒 βˆ’ 𝑖𝑓 ) ( 𝑔 βˆ’ π‘–β„Ž ) 𝐴2 + 𝐡^2= [( π‘Ž+ 𝑖𝑏 )(π‘Ž – 𝑖𝑏 )][(𝑐+ 𝑖𝑑)(𝑐 – 𝑖𝑑 )] [( 𝑒 + 𝑖𝑓) ( 𝑒 – 𝑖𝑓 )] [( 𝑔 + π‘–β„Ž ) ( 𝑔 – π‘–β„Ž)] π‘ˆπ‘ π‘–π‘›π‘” ( π‘₯ – 𝑦 ) ( π‘₯ + 𝑦 ) = π‘₯2+𝑦2 = [(π‘Ž)^2 – (𝑖𝑏)2] [ 𝑐2 – ( 𝑖𝑑)^2] [𝑒2βˆ’ (𝑖𝑓)^2 ] [𝑔2 – (βˆ’ π‘–β„Ž)]2 = [ π‘Ž2 βˆ’ 𝑏2 𝑖2 ] [ 𝑐2 βˆ’ 𝑖2 𝑑2 ] [ 𝑒2 βˆ’ 𝑖2 𝑓2 ] [ 𝑔2 βˆ’ 𝑖2 β„Ž2 ] Putting i2 = βˆ’1 = [ π‘Ž2– (βˆ’1)𝑏2] [ 𝑐2 – (βˆ’1) 𝑑 ] [ 𝑒2 – (βˆ’1) 𝑓)] [𝑔2 – (βˆ’1) β„Ž2 ] = [ π‘Ž2 + 𝑏2 ] [ 𝑐2 + 𝑑2 ] [ 𝑒2 + 𝑓2 ] [𝑔2 + β„Ž2 ] Hence, (π‘Ž2 + 𝑏2) (𝑐2 + 𝑑2) (𝑒2 + 𝑓2) (𝑔2 + β„Ž2) = 𝐴2 +𝐡2. 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 12 years. He provides courses for Maths and Science at Teachoo.