If z(1+a)=b+ic and a2+b2+c2=1 then (1+iz)(1−iz)=

1+iz1−iz=1+i(b+ic)/(1+a)1−i(b+ic)/(1+a)=1+a−c+ib1+a+c−ib=(1+a−c+ib)(1+a+c+ib)(1+a+c)2+b2=1+2a+a2−b2−c2+2ib+2iab1+a2+c2+b2+2ac+2(a+c)=2a+2a2+2ib+2iab2+2ac+2(a+c) ∵a2+b2+c2=1=a+a2+ib+iab1+ac+(a+c)=a(a+1)+ib(a+1)(a+1)(c+1)=a+ibc+1

If z(1+a)=b+ic and a2+b2+c2=1 then (1+iz)(1−iz)=

1+iz1−iz=1+i(b+ic)/(1+a)1−i(b+ic)/(1+a)=1+a−c+ib1+a+c−ib=(1+a−c+ib)(1+a+c+ib)(1+a+c)2+b2=1+2a+a2−b2−c2+2ib+2iab1+a2+c2+b2+2ac+2(a+c)=2a+2a2+2ib+2iab2+2ac+2(a+c) ∵a2+b2+c2=1=a+a2+ib+iab1+ac+(a+c)=a(a+1)+ib(a+1)(a+1)(c+1)=a+ibc+1

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

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$\frac{a + ib}{1 + c}$

$\frac{a + ic}{1 + b}$

none of these

$\frac{b - ic}{1 + a}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1 + iz}{1 - iz} = \frac{1 + i (b + ic) / (1 + a)}{1 - i (b + ic) / (1 + a)} \\ = \frac{1 + a - c + ib}{1 + a + c - ib} \\ = \frac{(1 + a - c + ib) (1 + a + c + ib)}{(1 + a + c)^{2} + b^{2}} \end{array}$
$\begin{array}{l} = \frac{1 + 2 a + a^{2} - b^{2} - c^{2} + 2 ib + 2 iab}{1 + a^{2} + c^{2} + b^{2} + 2 ac + 2 (a + c)} \\ = \frac{2 a + 2 a^{2} + 2 ib + 2 iab}{2 + 2 ac + 2 (a + c)} (∵ a^{2} + b^{2} + c^{2} = 1) \\ = \frac{a + a^{2} + ib + iab}{1 + ac + (a + c)} \\ = \frac{a (a + 1) + ib (a + 1)}{(a + 1) (c + 1)} = \frac{a + ib}{c + 1} \end{array}$