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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 +

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(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

Similar Questions

Let $z$ satisfy $\left| z \right| = 1$ and $z = 1 - \vec z$.

Statement $1$ : $z$ is a real number

Statement $2$ : Principal argument of $z$ is $\frac{\pi }{3}$

Let $z_1$ and $z_2$ be any two non-zero complex numbers such that $3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$. If $z = \frac{{3{z_1}}}{{2{z_2}}} + \frac{{2{z_2}}}{{3{z_1}}}$ then

The conjugate of complex number $\frac{{2 - 3i}}{{4 - i}},$ is

If for $z=\alpha+i \beta,|z+2|=z+4(1+i)$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation

The conjugate of the complex number $\frac{{2 + 5i}}{{4 - 3i}}$ is