If α and β are different complex numbers with ∣β∣=1 then ∣((β−α)/(1−αβ))∣=?


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Question Number 1331 by a@b.c last updated on 23/Jul/15

$If α and β are different complex numbers$ $with ∣ β ∣= 1 then ∣ \frac{β - α}{1 - α β} ∣= ?$
Commented by 123456 last updated on 23/Jul/15

$∣ α + β ∣⩽∣ α ∣ + ∣ β ∣$ $∣ β - α ∣⩽∣ β ∣ + ∣ α ∣$ $∣ α + (β - α) ∣⩽∣ α ∣ + ∣ β - α ∣$ $∣ β ∣⩽∣ α ∣ + ∣ β - α ∣$ $∣ β ∣ - ∣ α ∣⩽∣ β - α ∣⩽∣ β ∣ + ∣ α ∣$ $1 - ∣ α β ∣⩽∣ 1 - α β ∣⩽ 1 + ∣ α β ∣$ $∣ β ∣= 1 \Rightarrow β = e^{x i}, x \in R$ $1 - ∣ α ∣⩽∣ β - α ∣⩽ 1 + ∣ α ∣$ $1 - ∣ α ∣⩽∣ 1 - α β ∣⩽ 1 + ∣ α ∣$ $∣ \frac{β - α}{1 - α β} ∣= \frac{∣ β - α ∣}{∣ 1 - α β ∣}$ $α β \neq 1$ $α β \bar{β} \neq \bar{β}$ $α ∣ β ∣^{2} \neq \bar{β}$ $α \neq \bar{β}$ $∣ α ∣\neq∣ \bar{β} ∣=∣ β ∣= 1$ $∣ α ∣\neq 1$ $α \neq β \Rightarrow∣ β - α ∣\neq 0$
Commented by Rasheed Soomro last updated on 23/Jul/15
$The answer is α−dependant and it is also not fully free of β. It either depends upon Real(β) or Im(β). However the process of simplification may be as under: ∣ β ∣=1⇒( ∣β∣ )^2 =1⇒β β^− =1 Hence ∣ ((β−α)/(1−αβ)) ∣=∣ ((β−α)/(β β^− −α β)) ∣ (Replacing 1 by β β^− ) =∣ ((β−α)/(β ( β^− −α ))) ∣=∣ (1/β) ∣ . ∣ ((β−α)/(β^− −α)) ∣=(1/(∣ β ∣)) . ∣ ((β −α)/(β^− −α)) ∣ =∣ ((β−α)/(β^− −α)) ∣ ( ∣ β ∣=1 ) =∣ ((β−α)/(β^− −α)) −1+1 ∣=∣ ((β−α−β^− +α)/(β^− −α)) +1∣=∣ ((β−β^− )/(β^− −α))+1∣ Now β−β^− =2i[Re(β)] and β^− =−Re(β)+i[Im(β)^(−) =Re(β)−i[ Im(β) =∣ ((2i [Re(β)])/(Re(β)−i[ Im(β)−α)) +1∣ Here you can express Im(β) in terms of Re(β) or vice versa. As [ Re(β) ]^2 +[ Im(β) ]^2 =1 (∵ ∣β∣=1 ) So Im(β)=±(√(1−[ Re(β) ]^2 )) ∣ ((β−α)/(1−αβ)) ∣ = ∣ ((2i [Re(β)])/(Re(β)−i[ {±(√(1−[ Re(β) ]^2 ))}−α)) +1∣ Or ∣ ((β−α)/(1−αβ)) ∣=∣ ((2i {±(√(1−[Im(β)]^2 )) })/({±(√(1−[Im(β)]^2 )) }−i[ Im(β)−α)) +1∣$
$T h e a n s w e r i s α - d e p e n d a n t a n d i t i s a l s o n o t f u l l y f r e e o f β .$ $I t e i t h e r d e p e n d s u p o n R e a l (β) o r I m (β) .$ $H o w e v e r t h e p r o c e s s o f s i m p l i f i c a t i o n m a y b e a s u n d e r :$ $∣ β ∣= 1 \Rightarrow {(∣ β ∣)}^{2} = 1 \Rightarrow β \bar{β} = 1$ $H e n c e ∣ \frac{β - α}{1 - α β} ∣=∣ \frac{β - α}{β \bar{β} - α β} ∣ (R e p l a c i n g 1 b y β \bar{β})$ $=∣ \frac{β - α}{β (\bar{β} - α)} ∣=∣ \frac{1}{β} ∣ . ∣ \frac{β - α}{\bar{β} - α} ∣= \frac{1}{∣ β ∣} . ∣ \frac{β - α}{\bar{β} - α} ∣$ $=∣ \frac{β - α}{\bar{β} - α} ∣ (∣ β ∣= 1)$ $=∣ \frac{β - α}{\bar{β} - α} - 1 + 1 ∣=∣ \frac{β - α - \bar{β} + α}{\bar{β} - α} + 1 ∣=∣ \frac{β - \bar{β}}{\bar{β} - α} + 1 ∣$ $Extra \left or missing \right$ $=∣ \frac{2 i [R e (β)]}{R e (β) - i [I m (β) - α} + 1 ∣$ $H e r e y o u c a n e x p r e s s I m (β) i n t e r m s o f R e (β) o r v i c e v e r s a .$ $A s {[R e (β)]}^{2} + {[I m (β)]}^{2} = 1 (∵ ∣ β ∣= 1)$ $S o I m (β) = \pm \sqrt{1 - {[R e (β)]}^{2}}$ $∣ \frac{β - α}{1 - α β} ∣ = ∣ \frac{2 i [R e (β)]}{R e (β) - i [{\pm \sqrt{1 - {[R e (β)]}^{2}}} - α} + 1 ∣$ $O r$ $∣ \frac{β - α}{1 - α β} ∣=∣ \frac{2 i {\pm \sqrt{1 - {[I m (β)]}^{2}}}}{{\pm \sqrt{1 - {[I m (β)]}^{2}}} - i [I m (β) - α} + 1 ∣$
	Answered by prakash jain last updated on 23/Jul/15

	$∣ \frac{β - α}{1 - α β} ∣^{2} = (\frac{β - α}{1 - α β}) (\frac{\bar{β} - \bar{α}}{1 - \bar{α} \bar{β}}) = \frac{1 + ∣ α ∣^{2} - α \bar{β} - β \bar{α}}{1 + ∣ α ∣^{2} - α β - \bar{α} \bar{β}}$ $depends upon α and β .$ $The below variants of question can$ $give different answer .$ $∣ \frac{β - α}{1 - \bar{α} β} ∣^{2} = (\frac{β - α}{1 - \bar{α} β}) (\frac{\bar{β} - \bar{α}}{1 - α \bar{β}}) = \frac{1 + ∣ α ∣^{2} - α \bar{β} - β \bar{α}}{1 + ∣ α ∣^{2} - \bar{α} β - α \bar{β}} = 1$ $∣ \frac{β - α}{1 - α \bar{β}} ∣^{2} = (\frac{β - α}{1 - α \bar{β}}) (\frac{\bar{β} - \bar{α}}{1 - \bar{α} β}) = \frac{1 + ∣ α ∣^{2} - α \bar{β} - β \bar{α}}{1 + ∣ α ∣^{2} - \bar{α} β - α \bar{β}} = 1$
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