Misc 17 - Chapter 5 Class 11 If are different complex numbers - Proof- Using modulus conjugate property (Pg 102)

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Misc 17 If α and β are different complex numbers with |β| = 1, then find |(β − "α" )/(1 − 𝛼 ̅β)| . We know that |z|2 = (z) (𝑧 ̅) |(β−α)/(1−𝛼 ̅β)|^2=((β−α)/(1−𝛼 ̅β)) (((β−α)/(1−𝛼 ̅β)) ) ̅ = ((β−α)/(1−𝛼 ̅β)) ((β−α) ̅/(1−𝛼 ̅β) ̅ ) = ((β−α)/(1−𝛼 ̅β)) ((𝛽 ̅−𝛼 ̅)/(1 ̅−𝛼 ̿𝛽 ̅ )) = ((β−α)/(1−𝛼 ̅β)) ((𝛽 ̅−𝛼 ̅)/(1−𝛼𝛽 ̅ )) = ((𝛽 −𝛼)(𝛽 ̅ −𝛼 ̅))/((1−𝛼 ̅β)(1 − 𝛼 𝛽 ̅)) = (𝛽 (𝛽 ̅ − 𝛼 ̅ ) − 𝛼(𝛽 ̅ − 𝛼 ̅ ))/(1 (1− 𝛼 𝛽 ̅ ) − 𝛼 ̅𝛽 (1 − 𝛼𝛽 ̅ ) ) = (𝛽𝛽 ̅ − 𝛽𝛼 ̅ − 𝛼𝛽 ̅ + 𝛼𝛼 ̅)/(1 − 𝛼𝛽 ̅ − 𝛼 ̅𝛽 + 𝛼𝛼 ̅𝛽𝛽 ̅ ) = (|𝛽|^2 − 𝛽𝛼 ̅ − 𝛽 ̅𝛼 + |𝛼|^2)/(1 − 𝛼𝛽 ̅ − 𝛼 ̅𝛽 +|𝛼|^2 |𝛽|^2 ) Given that |β|=1 So, |β|^2=1 = (1− 𝛽𝛼 ̅ − 𝛽 ̅𝛼 + |𝛼|^2)/(1 − 𝛼𝛽 ̅ − 𝛼 ̅𝛽 + |𝛼|^2 1) = (1 − 𝛽𝛼 ̅ − 𝛽 ̅𝛼 + |𝛼|^2)/(1 − 𝛼𝛽 ̅ − 𝛼 ̅𝛽 + |𝛼|^2 ) = 1 Hence |(𝛽 − 𝛼)/(1 − 𝛼 ̅𝛽)|^2 = 1 |(𝛽 − 𝛼)/(1 −𝛼 ̅𝛽)| = √1 |(𝛽 − 𝛼)/(1 −𝛼 ̅𝛽)| =1

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