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

(A) $\left(2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8}\right)$

(B) $\left(2 \sec \frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$

(C) $\left(2 \sec \frac{5 \pi}{8}, \frac{3 \pi}{8}\right)$

(D) $\left(2 \sec \frac{11 \pi}{8}, \frac{11 \pi}{8}\right)$

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Complex Numbers

Let r and

θ

respectively be the modulus and amplitude of the complex number

z = 2 - i (2 tan \frac{5 π}{8})

, then

(r, θ)

is equal to

Answer: (A)

z = 2 - i (2 tan \frac{5 π}{8})

. Since

\frac{5 π}{8} = π - \frac{3 π}{8}

tan \frac{5 π}{8} = - tan \frac{3 π}{8}

.

So

z = 2 - i (- 2 tan \frac{3 π}{8}) = 2 + 2 i tan \frac{3 π}{8}

.

Both the real part

2

and the imaginary part

2 tan \frac{3 π}{8}

are positive (since

0 < \frac{3 π}{8} < \frac{π}{2}

), so

z

lies in the first quadrant.

r = 4 + 4 tan^{2} \frac{3 π}{8} = 2 1 + tan^{2} \frac{3 π}{8} = 2 sec \frac{3 π}{8}

(positive since

\frac{3 π}{8}

is in the first quadrant).

θ = tan^{- 1} (\frac{2 tan \frac{3 π}{8}}{2}) = \frac{3 π}{8}

(r, θ) = (2 sec \frac{3 π}{8}, \frac{3 π}{8})

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option