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NCERT Class XI Mathematics - Complex Numbers and Quadratic Equations - Solutions

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Question : 49 of 52
Marks: +1, -0
If α and β are different complex numbers with |β| = 1, then find ∣β−α1−α−β∣\left|\frac{\beta-\alpha}{1-\alpha^{-}\beta}\right|
Solution:  
We have , ∣β−α1−α−β∣2\left|\frac{\beta-\alpha}{1-\alpha^{-}\beta}\right|^2 = ∣β−α∣2∣1−α−β∣2\frac{|\beta-\alpha|^2}{|1-\alpha^{-}\beta|^2}
= (β−α)(β−α)‾(1−α−β)(1−α−β)‾\frac{(\beta-\alpha)\overline{(\beta-\alpha)}}{(1-\alpha^{-}\beta)\overline{(1-\alpha^{-}\beta)}} = (β−α)(β−−α−)(1−α−β)(1−αβ−)\frac{(\beta-\alpha)(\beta^{-}-\alpha^{-})}{(1-\alpha^{-}\beta)(1-\alpha\beta^{-})} = ββ−−βα−−αβ−+αα−1−αβ−−α−β+αα−ββ−\frac{\beta\beta^{-}-\beta\alpha^{-}-\alpha\beta^{-}+\alpha\alpha^{-}}{1-\alpha\beta^{-}-\alpha^{-}\beta+\alpha\alpha^{-}\beta\beta^{-}}
= 1−α−β−αβ−+αα−1−α−β−αβ−+α−α\frac{1-\alpha^{-}\beta-\alpha\beta^{-}+\alpha\alpha^{-}}{1-\alpha^{-}\beta-\alpha\beta^{-}+\alpha^{-}\alpha} = 1
∴ ∣β−α1−α−β∣2\left|\frac{\beta-\alpha}{1-\alpha^{-}\beta}\right|^2 = 1 ⇒ ∣β−α1−α−β∣\left|\frac{\beta-\alpha}{1-\alpha^{-}\beta}\right| = 1
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