Check sibling questions

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

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

This video is only available for Teachoo black users


Misc 11  - Chapter 5 Class 11 Complex Numbers - Part 4
Misc 11  - Chapter 5 Class 11 Complex Numbers - Part 5
Misc 11  - Chapter 5 Class 11 Complex Numbers - Part 6

This video is only available for Teachoo black users

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

Misc 11 (Method 1) If a + ib = (x + 𝑖)2/(2x^2 + 1) , prove that π‘Ž2 + 𝑏2 = (x^2+ 1)2/(2x^2+ 1)^2 π‘Ž + 𝑖𝑏 = (x + i)2/(2x2+ 1) Using ( π‘Ž + 𝑏 )^2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘ = (π‘₯2 + (𝑖)^2 + 2π‘₯𝑖)/(2π‘₯2+1) Putting 𝑖2 = βˆ’1 = (π‘₯2 βˆ’ 1 + 2π‘₯𝑖)/(2π‘₯2+ 1) = (x2 βˆ’ 1)/(2x2 + 1) + 𝑖 2x/(2x2 + 1) Hence π‘Ž + 𝑖𝑏 = (x2 βˆ’ 1)/(2x2 + 1) + 𝑖 2x/(2x2 + 1) Comparing real part π‘Ž = (π‘₯^2 βˆ’ 1)/(2π‘₯^2 + 1) Comparing imaginary part b = 2π‘₯/(2π‘₯2 + 1) Calculating π‘Ž2 + 𝑏2 π‘Ž2 + 𝑏2 = ((π‘₯^2 βˆ’ 1)/(2π‘₯2 + 1))^2 + (2π‘₯/(2π‘₯2 + 1))^2 = ((π‘₯2βˆ’ 1)2 + (2π‘₯)2)/((2π‘₯2 + 1)2) Using (π‘Ž βˆ’ 𝑏)^2 = π‘Ž2 + 𝑏2 βˆ’ 2π‘Žπ‘ = ((π‘₯2 )2 + (1)2 βˆ’ 2( π‘₯2)1 + 4π‘₯2)/( (2π‘₯2 + 1)2) = (π‘₯4 + 1 βˆ’2π‘₯2 + 4π‘₯2)/((2π‘₯2 +1)2) = (π‘₯4 + 1 + 2π‘₯2)/((2π‘₯2 + 1)2) = ((π‘₯2)2 + (1)2 + 2(π‘₯2) (1))/((2π‘₯^2 + 1)2) Using ( π‘Ž + 𝑏 )^2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘ = (π‘₯2+ 1)2/((2π‘₯2 + 1)2) Hence π‘Ž2 + 𝑏2 = (π‘₯2+ 1)2/((2π‘₯2 + 1)2) Hence proved Misc 11 (Method 2) If a + ib = (x + 𝑖)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Introduction (π‘Ž + 𝑖𝑏) ( π‘Ž – 𝑖𝑏) Using ( a – b ) ( a + b ) = a2 – b2 = π‘Ž2 – (𝑖𝑏)2 = π‘Ž2 – 𝑖2𝑏2 Putting i2 = βˆ’1 = π‘Ž2βˆ’ (βˆ’1) 𝑏2 = π‘Ž2 + 𝑏2 Hence, (π‘Ž + 𝑖𝑏) (π‘Ž – 𝑖𝑏) = π‘Ž2 + 𝑏2 Misc 11 (Method 2) If a + ib = (x + 𝑖)2/(2x^2 + 1) , prove that a2 + b2 = (x2 + 1)2/((2x2 + 1)2) Given π‘Ž + 𝑖𝑏 = (π‘₯ + 𝑖)2/(2π‘₯2 + 1) For π‘Ž – 𝑖𝑏 Replace 𝑖 by – 𝑖 in (1) π‘Ž – 𝑖𝑏 = (π‘₯ βˆ’ 𝑖)2/(2π‘₯2 + 1) Calculating (π‘Ž – 𝑖𝑏) (π‘Ž + 𝑖𝑏) (π‘Ž – 𝑖𝑏) (π‘Ž + 𝑖𝑏) = (π‘₯ βˆ’ 𝑖)2/(2π‘₯2 + 1) Γ— (π‘₯ + 𝑖)2/(2π‘₯2 + 1) π‘Ž2 + 𝑏2 = ((π‘₯ βˆ’ 𝑖)2 (π‘₯ + 𝑖)2)/(2π‘₯2 +1)2 = ( (π‘₯ βˆ’ 𝑖) (π‘₯ + 𝑖))^2/(2π‘₯2 +1)2 Using ( a – b ) ( a + b ) = a2 – b2 = (( π‘₯^2 βˆ’ (𝑖)^2 )^2 )/(2π‘₯^2 + 1)2 = γ€–( π‘₯2βˆ’ (βˆ’1)) γ€—^2/(2π‘₯2 + 1)2 = ( π‘₯2 + 1)2/(2π‘₯2 + 1)2 Hence a2 + b2 = (π‘₯2 + 1 )/(2π‘₯2 + 1) Hence proved

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

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.