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Misc 19 - Chapter 5 Class 11 Complex Numbers (Term 1)

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.