- Home
- Standard 11
- Mathematics
Let $\frac{{1 - ix}}{{1 + ix}} = a - ib$ and ${a^2} + {b^2} = 1$, where $a$ and $b$ are real, then $x = $
$\frac{{2a}}{{{{(1 + a)}^2} + {b^2}}}$
$\frac{{2b}}{{{{(1 + a)}^2} + {b^2}}}$
$\frac{{2a}}{{{{(1 + b)}^2} + {a^2}}}$
$\frac{{2b}}{{{{(1 + b)}^2} + {a^2}}}$
Solution
(b) $\frac{{1 – ix}}{{1 + ix}} = a – ib$
==> $\frac{{(1 – ix)(1 – ix)}}{{(1 + ix)(1 – ix)}} = a – ib$
==> $\frac{{1 – {x^2} – 2ix}}{{1 + {x^2}}} = a – ib$
==> $\frac{{1 – {x^2}}}{{1 + {x^2}}} = a$and $\frac{{2x}}{{1 + {x^2}}} = b$
Now we can write $x$ as $x = \frac{{\frac{{2x}}{{1 + {x^2}}}}}{{\frac{2}{{1 + {x^2}}}}} = \frac{{\frac{{2x}}{{1 + {x^2}}}}}{{\frac{{1 – {x^2}}}{{1 + {x^2}}} + 1}}$
$ = \frac{b}{{1 + a}} = \frac{{2b}}{{1 + 1 + 2a}} = \frac{{2b}}{{1 + ({a^2} + {b^2}) + 2a}} = \frac{{2b}}{{{{(1 + a)}^2} + {b^2}}}$
Trick : $\frac{{1 – ix}}{{1 + ix}} = \frac{{1 – {x^2} – 2ix}}{{1 + {x^2}}} = a – ib$
Let $a = 0$
==> $x = \pm 1$and $b = \pm 1$.
Also option $(b)$ gives $ \pm 1$.
