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

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

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

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

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

$\begin{bmatrix}\because a^{2}+b^{2}+c^{2}=1\\ \therefore b^{2}+c^{2}=1-a^{2}\end{bmatrix}$

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

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

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