If complex number z = x + iy is taken such th | vedclass

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Question

(d) $\frac{{z - 1}}{{z + 1}} = \frac{{(x + iy) - 1}}{{(x + iy) + 1}} = \frac{{(x - 1) + iy}}{{(x + 1) + iy}}$

$ = \,\frac{{\{ (x - 1) + iy\} \,\{ (

Vedclass · Accepted Answer

${x^2} + {y^2} - 2y = 1$

Vedclass · Answer

(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}}}} ight\}$$ + i\,\left\{ {\frac{{2y}}{{{{(x + 1)}^2} + {y^2}}}} ight\}$

$	herefore \,\,$$amp\left( {\frac{{z - 1}}{{z + 1}}} ight)$$ = {	an ^{ - 1}}\left\{ {\frac{{2y}}{{{{(x + 1)}^2} + {y^2}}} \div \frac{{({x^2} - 1) + {y^2}}}{{{{(x + 1)}^2} + {y^2}}}} ight\}$

==> $\frac{\pi }{4} = {	an ^{ - 1}}\left\{ {\frac{{2y}}{{{x^2} + {y^2} - 1}}} ight\}$==> $	an \frac{\pi }{4} = \frac{{2y}}{{{x^2} + {y^2} - 1}}$

==> $1 = \frac{{2y}}{{{x^2} + {y^2} - 1}}$

==> ${x^2} + {y^2} - 1 = 2y$

==> ${x^2} + {y^2} - 2y = 1$.

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

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