JEE Main 2024 — 29 January, Shift 2 — Question 6

$\mathrm{z}=2-\mathrm{i}\left(2 \tan \frac{5 \pi}{8}\right)=\mathrm{x}+\mathrm{iy}$ (let)

$r=\sqrt{x^{2}+y^{2}} \& \quad \theta=\tan ^{-1} \frac{y}{x}$

$r=\sqrt{(2)^{2}+\left(2 \tan \frac{5 \pi}{8}\right)^{2}}$

$=\left|2 \sec \frac{5 \pi}{8}\right|$

$=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right|$

$=2 \sec \frac{3 \pi}{8}$

And

$\theta=\tan ^{-1}\left(\frac{-2 \tan \frac{5 \pi}{8}}{2}\right)$

$=\tan ^{-1}\left(\tan \left(\pi-\frac{5 \pi}{8}\right)\right)$

$=\frac{3 \pi}{8}$

$(r, \theta)=\left(2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8}\right)$