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

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4-1.Complex numbers

medium

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

$0$

$ - \frac{1}{{|z + 1{|^2}}}$

$\left| {\frac{z}{{z + 1}}} \right|\,.\frac{1}{{|z + 1{|^2}}}$

$\frac{{\sqrt 2 }}{{|z + 1{|^2}}}$

(IIT-2003)

Solution

(a) $|z|\, = 1\, \Rightarrow \,|x + i\,y|\, = 1\, \Rightarrow \,{x^2} + {y^2} = 1$
$\omega = \frac{{z – 1}}{{z + 1}} = \frac{{(x – 1) + i\,y}}{{(x + 1) + i\,y}} \times \frac{{(x + 1) – i\,y}}{{(x + 1) – i\,y}}$
$ = \,\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)$
${\mathop{\rm Re}\nolimits} \,(\omega ) = 0$.

Standard 11

Mathematics

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normal

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easy

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$\frac{1}{1+i}$

medium

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normal

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hard

(JEE MAIN-2024)