If `z(1+a)=b+i ca n da^2+b^2+c^2=1,` then `[(1+i z)//(1-i z)=` `(a+i b)/(1+c)` b. `(b-i c)/(1+a)` c. `(a+i c)/(1+b)` d. none of these

Options

A. `(a+ib)/(1+c)`
B. `(b-ic)/(1+a)`
C. `(a+ic)/(1+b)`
D. none of these

Answer

A

Explanation

`(1+iz)/(1-izx)=(1+i(b+ic)//(1+a))/(1-i(b+ic)//(1+a))`
`=(1+a-c+ib)/(1+a+c-ib)`
`=((1+a-c+ib)(1+a+c+ib))/((1+a+c)^(2)+b^(2))`
`=(1+2a+a^(2)-b^(2)-c^(2)+2ib+2iab)/(1+a^(2)+c^(2)+b^(2)+2ac+2(a+c))`
`=(2a+2a^(2)+2ib+2iab)/(2+2ac+2(a+c))" "(therefore a^(2)+b^(2)+c^(2)=1)`
`=(a^(2)+a^(2)+ib+iab)/(1+ac+(a+c))`
`=(a(a+1)+ib(a+1))/((a+1)(c+1))=(a+ib)/(c+1)`