If z and omega are two non-zero complex numb | vedclass

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

Question

(d) $|z|\,|\omega |\, = 1$ ......$(i)$
and $arg\,\left( {\frac{z}{\omega }} \right) = \frac{\pi }{2}\,\,\, \Rightarrow \,\,\frac{z}{\omega } = i$ $&r

Vedclass · Accepted Answer

$-i$

Vedclass · Answer

(d) $|z|\,|\omega |\, = 1$ ......$(i)$
and $arg\,\left( {\frac{z}{\omega }} \right) = \frac{\pi }{2}\,\,\, \Rightarrow \,\,\frac{z}{\omega } = i$ $&rArr;$  $\left| {\frac{z}{\omega }} \right| = 1$ .....$(ii)$
From equation $(i)$ and $(ii)$
$|z|\, = \,|\omega |\, = 1$ and $\frac{z}{\omega } + \frac{{\bar z}}{{\bar \omega }} = 0;\,\,\,z\bar \omega + \bar z\omega = 0$
$\bar z\omega = - z\bar \omega = \frac{{ - z}}{\omega }\bar \omega \,\omega $; $\bar z\omega = - \,i\,|\omega {|^2} = - i.$.

If $z$ and $\omega $ are two non-zero complex numbers such that $|z\omega |\, = 1$ and $arg(z) - arg(\omega ) = \frac{\pi }{2},$ then $\bar z\omega $ is equal to

Similar Questions

The inequality $|z - 4|\, < \,|\,z - 2|$represents the region given by

Let $z$ =${i^{2i}}$ , then $|z|$ is (where $i$ =$\sqrt { - 1}$ )

If $\frac{{z - i}}{{z + i}}(z \ne - i)$ is a purely imaginary number, then $z.\bar z$ is equal to

If $arg\, z < 0$ then $arg\, (-z)\, -arg(z)$ is equal to

If the conjugate of $(x + iy)(1 - 2i)$ be $1 + i$, then