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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: if A and B are diffrent complex number with |B|=1 then find |B-A|/|1-'AB|
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neil_mahaseth (5)

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if A and B are diffrent complex number with |B|=1 then find |B-A|/|1-'AB|
 
 
'A=conjugate of A

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astatine19 (1354)

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write |B-A|/|1-'AB| as sqrt((B-A)('B-'A)/(1-'AB)(1-A'B)) expand, and substitute 'BB = |B|^2 = 1. It simplifies very easily.
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sinjan.j (574)

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|B| = 1
so |B|^2 = 1
or, BB' = 1
so, |B - A|/|1 - A'B|
= |B - A|/|BB' - A'B|
= |B - A|/(|B|.|B' - A'|)
now, |B - A| = |B' - A'|
so |B - A|/|1 - A'B|
= 1/|B|
= 1

btw..... the significance of B , A different is
|B - A | != 0


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sinjan.j (574)

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oops... I didnot see astatine19 's soln..... I posted my soln




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hsbhatt (5000)

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If |B| = 1 then BB' =1 or B' = 1/B
 
Now |1-A'B| = |1-AB'| (taking conjugate)
 
Hence, |B-A|/|1-AB'| = |B| |B-A|/|B-A| (Using B' = 1/B)
 
= |B| = 1

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