Question

If $a^2+b^2=1$, then $\frac{(1+b+ia)}{(1+b-ia)}$ is

• A. $1$
• B. $2$
• C. $b+ia$
• D. $a+ib$

Answer 1

The correct option is C $b+ia$

$a^2+b^2=1$

$⟹1-a^2=b^2$

$\frac{1+b+ia}{1+b-ia}$

$=\frac{1+b+ia}{1+b-ia}\times \frac{1+b+ia}{1+b+ia}$

$=\frac{(1+b{)}^2-a^2+2ia(1+b)}{(1+b{)}^2+a^2}$

$=\frac{b^2+2b+1-a^2+2ia(1+b)}{1+2b+b^2+a^2}$

$=\frac{2b^2+2b+2ia(1+b)}{2+2b}$

$=\frac{(2b+2ia)(1+b)}{2(1+b)}$

$=b+ia$

Answer-(C)