Question

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

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

Answer 1

The correct option is C $1+c$

Given
$a^2+b^2+c^2=1\dots \left(1\right)$
and $z=\frac{b+ic}{1+a}$
$\Rightarrow \frac{iz}{1}=\frac{ib-c}{1+a}$
Applying componendo and dividendo rule, we get
$\frac{1+iz}{1-iz}=\frac{1+a+ib-c}{1+a-ib+c}=\frac{(1+a-c)+ib}{(1+a+c)-ib}$
$=\frac{(1+a-c)+ib}{(1+a+c)-ib}\times \frac{(1+a+c)+ib}{(1+a+c)+ib}$
$=\frac{(1+a-c)(1+a+c)-b^2+ib(1+a-c+1+a+c)}{(1+a+c{)}^2+b^2}$
$=\frac{1+a^2+2a-c^2-b^2+2ib(1+a)}{1+a^2+b^2+c^2+2a+2c+2ac}$
$=\frac{a^2+2a+a^2+2ib(1+a)}{2+2a+2c+2ac}$
$[$from equation$\left(1\right)]$
$=\frac{2(1+a)(a+ib)}{2(1+a)(1+c)}$
$=\frac{a+ib}{1+c}$
On comparing $\frac{a+ib}{1+c}\text{with}\frac{a+ib}{k},$ we get,
$k=1+c$