Question
here ( |a+b i|=1 )
( operatorname{soa}^{wedge} 2+b^{wedge} 2=1 )
( (1+b+a i) /(1+b-a i) )
using concept of rationalization ( (1+b+a i)^{wedge} 2 /(1+b)^{wedge} 2+a^{wedge} 2 )
( =1+b^{wedge} 2+2 b-a^{wedge} 2+2 a i(1+b) /left(1+b^{wedge} 2+2 b+a^{wedge} 2right) )
using ( a^{wedge} 2+b^{wedge} 2=1 )
( 2 b^{wedge} 2+2 b+2 a i(1+b) / 2(1+b) )
( =(1+b)(b+a i) /(1+b) )
=b+ai

show that es of latībl=1 then It btai , btai – it ba i
Solution
