JEE Main 2019 — Complex Number Question with Solution

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

$\Rightarrow \theta -arg\left(\omega \right)=\frac{\mathrm{\pi }}{2}$

$\Rightarrow arg\left(\omega \right)=\theta -\frac{\mathrm{\pi }}{2}$

We know that, if $\left|z\right|=r \& arg\left(z\right)=\theta$ then $\frac{1}{\left|z\right|}=\frac{1}{r} \& arg\left(\overline{z}\right)=-arg\left(z\right)=-\theta ,$

Hence, $\left|\omega \right|=\frac{1}{r}$ and $arg\left(\overline{\omega }\right)=-arg\left(\omega \right)=\frac{\pi }{2}-\theta$

Now, by using Euler form i.e. if $\left|z\right|=r \& arg\left(z\right)=\theta ,$ then $z=r{e}^{i\theta },$ we have $z=r{e}^{i\theta }$ and $\omega =\frac{1}{r}{e}^{i(\theta -\frac{\pi }{2})}$

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

$\Rightarrow \overline{z}\omega =e^{-i\frac{\pi }{2}}$

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

$\overline{z}\omega =1\left[\cos \left(-\frac{\mathrm{\pi }}{2}\right)+i\sin \left(-\frac{\mathrm{\pi }}{2}\right)\right]=-i$

$\Rightarrow \overline{z}\omega =-i.$