JEE Main 2019 — Mathematics Algebra

Let $|z|=r$∣z∣=r and $arg(z)=\theta ,$arg(z)=θ, given $arg(z)-arg(\omega )=\frac{\pi }{2}$arg(z)−arg(ω)=2π​

$$\Rightarrow \theta -arg(\omega )=\frac{\pi }{2}$$⇒θ−arg(ω)=2π​

$$\Rightarrow arg(\omega )=\theta -\frac{\pi }{2}$$⇒arg(ω)=θ−2π​

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

Hence, $|\omega |=\frac{1}{r}$∣ω∣=r1​ and $arg(\bar{\omega })=-arg(\omega )=\frac{\pi }{2}-\theta$arg(ωˉ)=−arg(ω)=2π​−θ

Now, by using Euler form i.e. if |z|=r&arg(z)=\theta , then $z=r{e}^{i\theta },$z=reiθ, we have $z=r{e}^{i\theta }$z=reiθ and $\omega =\frac{1}{r}{e}^{i(\theta -\frac{\pi }{2})}$ω=r1​ei(θ−2π​)

Also, $\bar{z}\omega =r {e}^{-i\theta }\times (\frac{1}{r} {e}^{i(\theta -\frac{\pi }{2})})$zˉω=re−iθ×(r1​ei(θ−2π​))

$$\Rightarrow \bar{z}\omega ={e}^{-i\frac{\pi }{2}}$$⇒zˉω=e−i2π​

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

$$\bar{z}\omega =1[\mathrm{cos}(-\frac{\pi }{2})+i\mathrm{sin}(-\frac{\pi }{2})]=-i$$zˉω=1[cos(−2π​)+isin(−2π​)]=−i

$$\Rightarrow \bar{z}\omega =-i.$$⇒zˉω=−i.