If z and ω are two complex numbers such that |zω| = 1 and arg(z) -...

Let $\left|z\right| = r$ and $arg\left(z\right) = \theta,$ given $arg\left(z\right) - arg\left(\omega\right) = \dfrac{\pi}{2}$

$⇒ \theta - arg\left(\omega\right) = \dfrac{\pi}{2}$

$⇒ arg\left(\omega\right) = \theta - \dfrac{\pi}{2}$

We know that, if \left|z\right| = r \text{ and } arg\left(z\right) = \theta then \dfrac{1}{\left|z\right|} = \dfrac{1}{r} \text{ and } arg\left(\overset{¯}{z}\right) = - arg\left(z\right) = - \theta,

Hence, \left|\omega\right| = \dfrac{1}{r}$and$arg( \overset{¯}{\omega} ) = - arg\left(\omega\right) = \dfrac{\pi}{2} - \theta$</p><p>Now, by using Euler form i.e. if$\left|z\right| = r \text{$and$} arg\left(z\right) = \theta,$then$z = r{e}^{i\theta},$we have$z = r{e}^{i\theta} and $\omega = \dfrac{1}{r}{e}^{i ( \theta - \dfrac{\pi}{2} )}$

Also, $\overset{¯}{z}\omega = r{e}^{- i\theta} \times \left(\dfrac{1}{r}{e}^{i\left(\theta - \dfrac{\pi}{2}\right)}\right)$

$⇒ \overset{¯}{z}\omega = {e}^{- i\dfrac{\pi}{2}}$

Using polar form i.e. ${e}^{i\theta} = cos\theta + isin\theta,$ we get

$\overset{¯}{z}\omega = 1\left[\cos \left(- \dfrac{\pi}{2}\right) + isin\left(- \dfrac{\pi}{2}\right)\right] = - i$

$⇒ \overset{¯}{z}\omega = - i .</p>$