Question

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

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

Answer 1

The correct option is C

$b+ia$

To find the value of $\frac{(1+b+ia)}{(1+b-ia)}$:

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

Multiply numerator and denominator with $(1+b+ia)$

$$\begin{gathered} =\frac{(1+b+ia)}{(1+b-ia)}\frac{(1+b+ia)}{(1+b+ia)} \\ =\frac{{(1+b+ia)}^2}{{(1+b)}^2-{\left(ia\right)}^2} \\ =\frac{1+b^2-a^2+2b+i2ab+i2a}{1+b^2+2b+a^2} \end{gathered}$$

$=\frac{a^2+b^2+b^2-a^2+2b+i2ab+i2a}{1+1+2b}$ $($ since $a^2+b^2=1)$

$$\begin{gathered} =\frac{2b^2+2b+i2ab+i2a}{2+2b} \\ =\frac{2b(b+1)+2ia(b+1)}{2(b+1)} \\ =b+ia \end{gathered}$$

Hence the correct option is C.