4-1.Complex numbers
medium

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

Solution

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

Standard 11
Mathematics

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