Question

If $b+ic=(1+a)z$ and $a^2+b^2+c^2$ = 1, then $\frac{1+iz}{1-iz}$ is equal to

• A. $\frac{a-ib}{1-c}$
• B. $\frac{a-ib}{1+c}$
• C. $\frac{a+ib}{1+c}$
• D. $\frac{a+ib}{1-c}$

Answer 1

Answer: (C)

Since $z=\frac{b+ic}{1+a}$
$\therefore iz=\frac{ib-c}{1+a}$
$\therefore \frac{1+iz}{1-iz}=\frac{1+\frac{ib-c}{1+a}}{1-\frac{ib-c}{1+a}}$
$=\frac{1+a+ib-c}{1+a-ib+c}$

$=\frac{\left(1+a-c\right)+ib}{\left(1+a+c\right)-ib}\cdot \frac{\left(1+a+c\right)+ib}{\left(1+a+c\right)+ib}$

$=\frac{{\left(1+a+ib\right)}^2-c^2}{{\left(1+a+c\right)}^2+b^2}$

$=\frac{1+a^2-b^2-c^2+2a+2ib+2iab}{1+a^2+c^2+b^2+2a+2c+2ac}$

$=\frac{1+a^2-\left(1-a^2\right)+2a+2ib+2iab}{1+1+2a+2c+2ac}$

$\left[\begin{array}{c}\because a^2+b^2+c^2=1\\ \therefore b^2+c^2=1-a^2\end{array}\right]$

$=\frac{2a^2+2a+2ib+2iab}{2+2a+2c+2ac}$

$=\frac{2a\left(a+1\right)+2ib\left(a+1\right)}{2\left(a+1\right)+2c\left(a+1\right)}$

$=\frac{2\left(a+ib\right)+\left(a+1\right)}{2\left(a+1\right)\left(1+c\right)}=\frac{a+ib}{1+c}$