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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 ∣((β−α)/(1−αβ))∣=?
Ifαandβaredifferentcomplexnumberswith∣β∣=1then∣β−α1−αβ∣=?
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−∣αβ∣≤∣1−αβ∣≤1+∣αβ∣ ∣β∣=1⇒β=e^(xi) ,x∈R 1−∣α∣≤∣β−α∣≤1+∣α∣ 1−∣α∣≤∣1−αβ∣≤1+∣α∣ ∣((β−α)/(1−αβ))∣=((∣β−α∣)/(∣1−αβ∣)) αβ≠1 αββ^� ≠β^� α∣β∣^2 ≠β^� α≠β^� ∣α∣≠∣β^� ∣=∣β∣=1 ∣α∣≠1 α≠β⇒∣β−α∣≠0
∣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−∣αβ∣⩽∣1−αβ∣⩽1+∣αβ∣∣β∣=1⇒β=exi,x∈R1−∣α∣⩽∣β−α∣⩽1+∣α∣1−∣α∣⩽∣1−αβ∣⩽1+∣α∣∣β−α1−αβ∣=∣β−α∣∣1−αβ∣αβ≠1αββ¯≠β¯α∣β∣2≠β¯α≠β¯∣α∣≠∣β¯∣=∣β∣=1∣α∣≠1α≠β⇒∣β−α∣≠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∣
Theanswerisα−dependantanditisalsonotfullyfreeofβ.IteitherdependsuponReal(β)orIm(β).Howevertheprocessofsimplificationmaybeasunder:∣β∣=1⇒(∣β∣)2=1⇒ββ−=1Hence∣β−α1−αβ∣=∣β−αββ−−αβ∣(Replacing1byββ−)=∣β−αβ(β−−α)∣=∣1β∣.∣β−αβ−−α∣=1∣β∣.∣β−αβ−−α∣=∣β−αβ−−α∣(∣β∣=1)=∣β−αβ−−α−1+1∣=∣β−α−β−+αβ−−α+1∣=∣β−β−β−−α+1∣Extra \left or missing \right=∣2i[Re(β)]Re(β)−i[Im(β)−α+1∣HereyoucanexpressIm(β)intermsofRe(β)orviceversa.As[Re(β)]2+[Im(β)]2=1(∵∣β∣=1)SoIm(β)=±1−[Re(β)]2∣β−α1−αβ∣=∣2i[Re(β)]Re(β)−i[{±1−[Re(β)]2}−α+1∣Or∣β−α1−αβ∣=∣2i{±1−[Im(β)]2}{±1−[Im(β)]2}−i[Im(β)−α+1∣
Answered by prakash jain last updated on 23/Jul/15
∣((β−α)/(1−αβ))∣^2 =(((β−α)/(1−αβ)))(((β^− −α^− )/(1−α^− β^− )))=((1+∣α∣^2 −αβ^− −βα^− )/(1+∣α∣^2 −αβ−α^− β^− )) depends upon α and β. The below variants of question can give different answer. ∣((β−α)/(1−α^− β))∣^2 =(((β−α)/(1−α^− β)))(((β^− −α^− )/(1−αβ^− )))=((1+∣α∣^2 −αβ^− −βα^− )/(1+∣α∣^2 −α^− β−αβ^− ))=1 ∣((β−α)/(1−αβ^− ))∣^2 =(((β−α)/(1−αβ^− )))(((β^− −α^− )/(1−α^− β)))=((1+∣α∣^2 −αβ^− −βα^− )/(1+∣α∣^2 −α^− β−αβ^− ))=1
∣β−α1−αβ∣2=(β−α1−αβ)(β−−α−1−α−β−)=1+∣α∣2−αβ−−βα−1+∣α∣2−αβ−α−β−dependsuponαandβ.Thebelowvariantsofquestioncangivedifferentanswer.∣β−α1−α−β∣2=(β−α1−α−β)(β−−α−1−αβ−)=1+∣α∣2−αβ−−βα−1+∣α∣2−α−β−αβ−=1∣β−α1−αβ−∣2=(β−α1−αβ−)(β−−α−1−α−β)=1+∣α∣2−αβ−−βα−1+∣α∣2−α−β−αβ−=1

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