If |z|, = 1 and omega = frac{{z - 1}}{{z + 1} | vedclass

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(a) $|z|\, = 1\, \Rightarrow \,|x + i\,y|\, = 1\, \Rightarrow \,{x^2} + {y^2} = 1$
$\omega = \frac{{z - 1}}{{z + 1}} = \frac{{(x - 1) + i\,y}}{{(x +

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(a) $|z|\, = 1\, \Rightarrow \,|x + i\,y|\, = 1\, \Rightarrow \,{x^2} + {y^2} = 1$
$\omega = \frac{{z - 1}}{{z + 1}} = \frac{{(x - 1) + i\,y}}{{(x + 1) + i\,y}} 	imes \frac{{(x + 1) - i\,y}}{{(x + 1) - i\,y}}$
$ = \,\frac{{({x^2} + {y^2} - 1)}}{{{{(x + 1)}^2} + {y^2}}} + \frac{{2i\,y}}{{{{(x + 1)}^2} + {y^2}}} = \frac{{2i\,y}}{{{{(x + 1)}^2} + {y^2}}}$$(\because \,{x^2} + {y^2} = 1)$
${\mathop{m Re}
olimits} \,(\omega ) = 0$.

If $|z|\, = 1$ and $\omega = \frac{{z - 1}}{{z + 1}}$ (where $z \ne - 1)$, then ${\mathop{\rm Re}\nolimits} (\omega )$ is

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