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

Solution:

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

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

Solution:

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