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If complex number $z = x + iy$ is taken such that the amplitude of fraction $\frac{{z - 1}}{{z + 1}}$ is always $\frac{\pi }{4}$, then
${x^2} + {y^2} + 2y = 1$
${x^2} + {y^2} - 2y = 0$
${x^2} + {y^2} + 2y = - 1$
${x^2} + {y^2} - 2y = 1$
Solution
(d) $\frac{{z – 1}}{{z + 1}} = \frac{{(x + iy) – 1}}{{(x + iy) + 1}} = \frac{{(x – 1) + iy}}{{(x + 1) + iy}}$
$ = \,\frac{{\{ (x – 1) + iy\} \,\{ (x + 1) – iy\} }}{{\{ (x + 1) + iy\} \,\{ (x + 1) – iy\} }}$
$ = \frac{{\{ ({x^2} – 1) + {y^2}\} + i\{ y(x + 1) – y(x – 1)\} }}{{{{(x + 1)}^2} + {y^2}}}$
$ = \left\{ {\frac{{({x^2} – 1) + {y^2}}}{{{{(x + 1)}^2} + {y^2}}}} \right\}$$ + i\,\left\{ {\frac{{2y}}{{{{(x + 1)}^2} + {y^2}}}} \right\}$
$\therefore \,\,$$amp\left( {\frac{{z – 1}}{{z + 1}}} \right)$$ = {\tan ^{ – 1}}\left\{ {\frac{{2y}}{{{{(x + 1)}^2} + {y^2}}} \div \frac{{({x^2} – 1) + {y^2}}}{{{{(x + 1)}^2} + {y^2}}}} \right\}$
==> $\frac{\pi }{4} = {\tan ^{ – 1}}\left\{ {\frac{{2y}}{{{x^2} + {y^2} – 1}}} \right\}$==> $\tan \frac{\pi }{4} = \frac{{2y}}{{{x^2} + {y^2} – 1}}$
==> $1 = \frac{{2y}}{{{x^2} + {y^2} – 1}}$
==> ${x^2} + {y^2} – 1 = 2y$
==> ${x^2} + {y^2} – 2y = 1$.
