Question

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

• A. $\frac{a+ib}{1+c}$
• B. $\frac{b-ic}{1+a}$
• C. $\frac{a+ic}{1+b}$
• D. None of these

Answer 1

The correct option is A $\frac{a+ib}{1+c}$

$\frac{1+iz}{1-iz}=\frac{1+i(b+ic)/(1+a)}{1-i(b+ic)/(1+a)}$
$=\frac{1+a-c+ib}{1+a+c-ib}$
$=\frac{(1+a-c+ib)(1+a+c+ib)}{(1+a+c{)}^2+b^2}$
$=\frac{1+2a+a^2-b^2-c^2+2ib+2iab}{1+a^2+c^2+b^2+2ac+2(a+c)}$
$=\frac{2a+2a^2+2ib+2iab}{2+2ac+2(a+c)}(\because a^2+b^2+c^2=1)$
$=\frac{a+a^2+ib+iab}{1+ac+(a+c)}$
$=\frac{a(a+1)+ib(a+1)}{(a+1)(c+1)}$
$=\frac{a+ib}{c+1}$