[UPLIFTJEE]

Let r and $\theta$ respectively be the modulus and amplitude of the complex number $\mathrm{z}=2-\mathrm{i}\left(2 \tan \frac{5 \pi}{8}\right)$, then $(\mathrm{r}, \theta)$ is equal to
[JEE (Main) 29 Jan 2024 S II]
(A) $\left(2 \sec \frac{11 \pi}{8}, \frac{11 \pi}{8}\right)$
(B) $\left(2 \sec \frac{5 \pi}{8}, \frac{3 \pi}{8}\right)$
(C) $\left(2 \sec \frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$
(D) $\left(2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8}\right)$

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[UPLIFTJEE]

Solution

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

\[ \begin{aligned} & |z|=r=\sqrt{x^{2}+y^{2}}=\sqrt{(2)^{2}+\left(2 \tan \frac{5 \pi}{8}\right)^{2}}=\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right|=2 \sec \frac{3 \pi}{8}\left(\sec \frac{5 \pi}{8}<0\right) \\ & \arg (z)=\theta=\tan ^{-1}\left(\frac{y}{x}\right) \text { as } x, y>0=\tan ^{-1}\left(\frac{-2 \tan \frac{5 \pi}{8}}{2}\right)=\tan ^{-1}\left(\tan \left(\pi-\frac{5 \pi}{8}\right)\right)=\tan ^{-1}\left(\tan \frac{3 \pi}{8}\right)=\frac{3 \pi}{8} \end{aligned} \]

[UPLIFTJEE]

Answer

D

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