If a^2+b^2=1t h e n(1+b+i a)/(1+b-i a)= ...

If `a^2+b^2=1t h e n(1+b+i a)/(1+b-i a)=` `1` b. `2` c. `b+i a` d. `a+i b`

a + ib

a + ia

b+ ia

b + ib

Text Solution

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The correct Answer is:

Given that `a^(2) +b^(2)=1` Therefore,
`(1+b + ia)/(1+b -ia) =((1+b+ia)(a+b+ia))/((1+ b-ia)(1+b+ia))`
`= ((1+b)^(2) -a^(2)+2ia(1+b))/(1+ b^(2)+2b+a^(2))`
`= ((1-a^(2)) + 2b b^(2) + 2ia(1+b))/(2(l+b))`
`= (2b^(2) + 2b + 2ia(1+b))/(2(1+b))= b + ia`

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