Misc 11 - If a + ib = (x + i)2/(2x2 + 1), prove a2 + b2 - Miscellaneous


  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
Ask Download

Transcript

Misc 11 (Method 1) If a + ib = (x + ๐‘–)2/(2x^2 + 1) , prove that ๐‘Ž2 + ๐‘2 = (x^2+ 1)2/(2x^2+ 1)^2 ๐‘Ž + ๐‘–๐‘ = (x + i)2/(2x2+ 1) Using ( ๐‘Ž + ๐‘ )^2 = ๐‘Ž2 + ๐‘2 + 2๐‘Ž๐‘ = (๐‘ฅ2 + (๐‘–)^2 + 2๐‘ฅ๐‘–)/(2๐‘ฅ2+1) Putting ๐‘–2 = โˆ’1 = (๐‘ฅ2 โˆ’ 1 + 2๐‘ฅ๐‘–)/(2๐‘ฅ2+ 1) = (x2 โˆ’ 1)/(2x2 + 1) + ๐‘– 2x/(2x2 + 1) Hence ๐‘Ž + ๐‘–๐‘ = (x2 โˆ’ 1)/(2x2 + 1) + ๐‘– 2x/(2x2 + 1) Comparing real part ๐‘Ž = (๐‘ฅ^2 โˆ’ 1)/(2๐‘ฅ^2 + 1) Comparing imaginary part b = 2๐‘ฅ/(2๐‘ฅ2 + 1) Calculating ๐‘Ž2 + ๐‘2 ๐‘Ž2 + ๐‘2 = ((๐‘ฅ^2 โˆ’ 1)/(2๐‘ฅ2 + 1))^2 + (2๐‘ฅ/(2๐‘ฅ2 + 1))^2 = ((๐‘ฅ2โˆ’ 1)2 + (2๐‘ฅ)2)/((2๐‘ฅ2 + 1)2) Using (๐‘Ž โˆ’ ๐‘)^2 = ๐‘Ž2 + ๐‘2 โˆ’ 2๐‘Ž๐‘ = ((๐‘ฅ2 )2 + (1)2 โˆ’ 2( ๐‘ฅ2)1 + 4๐‘ฅ2)/( (2๐‘ฅ2 + 1)2) = (๐‘ฅ4 + 1 โˆ’2๐‘ฅ2 + 4๐‘ฅ2)/((2๐‘ฅ2 +1)2) = (๐‘ฅ4 + 1 + 2๐‘ฅ2)/((2๐‘ฅ2 + 1)2) = ((๐‘ฅ2)2 + (1)2 + 2(๐‘ฅ2) (1))/((2๐‘ฅ^2 + 1)2) Using ( ๐‘Ž + ๐‘ )^2 = ๐‘Ž2 + ๐‘2 + 2๐‘Ž๐‘ = (๐‘ฅ2+ 1)2/((2๐‘ฅ2 + 1)2) Hence ๐‘Ž2 + ๐‘2 = (๐‘ฅ2+ 1)2/((2๐‘ฅ2 + 1)2) Hence proved Misc 11 (Method 2) If a + ib = (x + ๐‘–)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Introduction (๐‘Ž + ๐‘–๐‘) ( ๐‘Ž โ€“ ๐‘–๐‘) Using ( a โ€“ b ) ( a + b ) = a2 โ€“ b2 = ๐‘Ž2 โ€“ (๐‘–๐‘)2 = ๐‘Ž2 โ€“ ๐‘–2๐‘2 Putting i2 = โˆ’1 = ๐‘Ž2โˆ’ (โˆ’1) ๐‘2 = ๐‘Ž2 + ๐‘2 Hence, (๐‘Ž + ๐‘–๐‘) (๐‘Ž โ€“ ๐‘–๐‘) = ๐‘Ž2 + ๐‘2 Misc 11 (Method 2) If a + ib = (x + ๐‘–)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Given ๐‘Ž + ๐‘–๐‘ = (๐‘ฅ + ๐‘–)2/(2๐‘ฅ2 + 1) For ๐‘Ž โ€“ ๐‘–๐‘ Replace ๐‘– by โ€“ ๐‘– in (1) ๐‘Ž โ€“ ๐‘–๐‘ = (๐‘ฅ โˆ’ ๐‘–)2/(2๐‘ฅ2 + 1) Calculating (๐‘Ž โ€“ ๐‘–๐‘) (๐‘Ž + ๐‘–๐‘) (๐‘Ž โ€“ ๐‘–๐‘) (๐‘Ž + ๐‘–๐‘) = (๐‘ฅ โˆ’ ๐‘–)2/(2๐‘ฅ2 + 1) ร— (๐‘ฅ + ๐‘–)2/(2๐‘ฅ2 + 1) ๐‘Ž2 + ๐‘2 = ((๐‘ฅ โˆ’ ๐‘–)2 (๐‘ฅ + ๐‘–)2)/(2๐‘ฅ2 +1)2 = ( (๐‘ฅ โˆ’ ๐‘–) (๐‘ฅ + ๐‘–))^2/(2๐‘ฅ2 +1)2 Using ( a โ€“ b ) ( a + b ) = a2 โ€“ b2 = (( ๐‘ฅ^2 โˆ’ (๐‘–)^2 )^2 )/(2๐‘ฅ^2 + 1)2 = ใ€–( ๐‘ฅ2โˆ’ (โˆ’1)) ใ€—^2/(2๐‘ฅ2 + 1)2 = ( ๐‘ฅ2 + 1)2/(2๐‘ฅ2 + 1)2 Hence a2 + b2 = (๐‘ฅ2 + 1 )/(2๐‘ฅ2 + 1) Hence proved

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.