1
Question
If α and β are different complex numbers with |α|=1, then what is ∣∣
∣∣α−β1−α¯¯¯β∣∣
∣∣ equal to?
If α and β are different complex numbers with |α|=1, then what is ∣∣
∣∣α−β1−α¯¯¯β∣∣
∣∣ equal to?
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Solution
The correct option is C 1
|α|=1, we know that |z|2=(z)(¯¯¯z)∣∣
∣∣α−β1−α¯¯¯β∣∣
∣∣2=(α−β1−α¯¯¯β)(¯¯¯¯¯¯¯¯¯¯¯¯¯α−β1−α¯¯¯β)|α|=1∴α¯¯¯¯α=1=(α−β1−α¯¯¯β)(¯¯¯¯¯¯¯¯¯¯¯¯¯α−β1−α¯¯¯β)=(α−β1−α¯¯¯β)(¯¯¯¯α−¯¯¯β1−¯¯¯¯α¯¯¯β)(¯¯¯x=x) [¯¯¯1=1]=(α−β1−α¯¯¯β)(¯¯¯¯α−¯¯¯β1−¯¯¯¯αβ)=(α−β)(¯¯¯¯α−¯¯¯β)(1−α¯¯¯β)(1−¯¯¯¯αβ)=α¯¯¯¯α−α¯¯¯β−β¯¯¯¯α+β¯¯¯β1−¯¯¯¯αβ−α¯¯¯β+α¯¯¯¯αβ¯¯¯βSubstitute α¯¯¯¯α=1→1−α¯¯¯β−β¯¯¯¯α+β¯¯¯β1−¯¯¯¯αβ−α¯¯¯β+β¯¯¯β=1∴∣∣∣α−β1−α¯¯¯β∣∣∣=1.
|α|=1, we know that |z|2=(z)(¯¯¯z)
∣∣
∣∣α−β1−α¯¯¯β∣∣
∣∣2=(α−β1−α¯¯¯β)(¯¯¯¯¯¯¯¯¯¯¯¯¯α−β1−α¯¯¯β)|α|=1∴α¯¯¯¯α=1
=(α−β1−α¯¯¯β)(¯¯¯¯¯¯¯¯¯¯¯¯¯α−β1−α¯¯¯β)=(α−β1−α¯¯¯β)(¯¯¯¯α−¯¯¯β1−¯¯¯¯α¯¯¯β)
(¯¯¯x=x)
[¯¯¯1=1]=(α−β1−α¯¯¯β)(¯¯¯¯α−¯¯¯β1−¯¯¯¯αβ)=(α−β)(¯¯¯¯α−¯¯¯β)(1−α¯¯¯β)(1−¯¯¯¯αβ)
=α¯¯¯¯α−α¯¯¯β−β¯¯¯¯α+β¯¯¯β1−¯¯¯¯αβ−α¯¯¯β+α¯¯¯¯αβ¯¯¯β
Substitute α¯¯¯¯α=1→1−α¯¯¯β−β¯¯¯¯α+β¯¯¯β1−¯¯¯¯αβ−α¯¯¯β+β¯¯¯β=1
∴∣∣∣α−β1−α¯¯¯β∣∣∣=1.
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