- Home
- Standard 11
- Mathematics
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
A
$0$
B
$ - \frac{1}{{|z + 1{|^2}}}$
C
$\left| {\frac{z}{{z + 1}}} \right|\,.\frac{1}{{|z + 1{|^2}}}$
D
$\frac{{\sqrt 2 }}{{|z + 1{|^2}}}$
(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
