JEE Main 2021 — Mathematics Algebra

As$|z\omega |=1$∣zω∣=1

If $|z|=r$∣z∣=r, then $|\omega |=\frac{1}{r}$∣ω∣=r1​

Let $\mathrm{arg}(z)=\theta$arg(z)=θ

$$\therefore \mathrm{arg}(\omega )=(\theta -\frac{3\pi }{2})$$∴arg(ω)=(θ−23π​)

So,

$$z=r{e}^{i\theta }$$z=reiθ

$$\Rightarrow \bar{z}=r{e}^{i(-\theta )}$$⇒zˉ=rei(−θ)

$$\Rightarrow \omega =\frac{1}{r}{e}^{i(\theta -\frac{3\pi }{2})}$$⇒ω=r1​ei(θ−23π​)

Now, consider

$$\frac{1-2\bar{z}\omega }{1+3\bar{z}\omega }=\frac{1-2{e}^{i(-\frac{3\pi }{2})}}{1+3{e}^{i(-\frac{3\pi }{2})}}=(\frac{1-2i}{1+3i})$$1+3zˉω1−2zˉω​=1+3ei(−23π​)1−2ei(−23π​)​=(1+3i1−2i​)

$$=\frac{(1-2i)(1-3i)}{(1+3i)(1-3i)}=\frac{1-5i+6{i}^{2}}{10}=\frac{-5+5i}{10}=-\frac{1}{2}(1+i)$$=(1+3i)(1−3i)(1−2i)(1−3i)​=101−5i+6i2​=10−5+5i​=−21​(1+i)

Then,

$$\mathrm{principal}\mathrm{arg}(\frac{1-2\bar{z}\omega }{1+3\bar{z}\omega })$$principalarg(1+3zˉω1−2zˉω​)

$$=\mathrm{principal}\mathrm{arg}(-\frac{1}{2}(1+i))$$=principalarg(−21​(1+i))

$$=-(\pi -\frac{\pi }{4})=\frac{-3\pi }{4}$$=−(π−4π​)=4−3π​