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Mathematics Question on complex numbers

Let rr and θ\theta respectively be the modulus and amplitude of the complex number z=2−i(2tan⁡5π8)z = 2 - i \left( 2 \tan \frac{5\pi}{8} \right), then (r,θ)(r, \theta) is equal to

A

(2sec⁡3π8,3π8)\left( 2 \sec \frac{3\pi}{8}, \frac{3\pi}{8} \right)

B

(2sec⁡3π8,5π8)\left( 2 \sec \frac{3\pi}{8}, \frac{5\pi}{8} \right)

C

(2sec⁡5π8,3π8)\left( 2 \sec \frac{5\pi}{8}, \frac{3\pi}{8} \right)

D

(2sec⁡11π8,11π8)\left( 2 \sec \frac{11\pi}{8}, \frac{11\pi}{8} \right)

Answer

(2sec⁡3π8,3π8)\left( 2 \sec \frac{3\pi}{8}, \frac{3\pi}{8} \right)

Explanation

Solution

Let z=2−i(2tan⁡5π8)=x+iyz = 2 - i \left( 2 \tan \frac{5\pi}{8} \right) = x + iy.

Step 1. Calculating rr, the modulus of zz:**
r=x2+y2r = \sqrt{x^2 + y^2}
r=(2)2+(2tan⁡5π8)2r = \sqrt{(2)^2 + \left( 2 \tan \frac{5\pi}{8} \right)^2}
=2sec⁡5π8=2sec⁡(π−3π8)=2sec⁡3π8= 2 \sec \frac{5\pi}{8} = 2 \sec \left( \pi - \frac{3\pi}{8} \right) = 2 \sec \frac{3\pi}{8}

Step 2. Calculating θ\theta, the amplitude of zz:
θ=tan⁡−1(−2tan⁡5π82)\theta = \tan^{-1} \left( \frac{-2 \tan \frac{5\pi}{8}}{2} \right)
=tan⁡−1(−tan⁡5π8)=3π8= \tan^{-1} \left( -\tan \frac{5\pi}{8} \right) = \frac{3\pi}{8}
Therefore, (r,θ)=(2sec⁡3π8,3π8)(r, \theta) = \left( 2 \sec \frac{3\pi}{8}, \frac{3\pi}{8} \right).

The Correct Answer is:(2sec⁡3π8,3π8)\left( 2 \sec \frac{3\pi}{8}, \frac{3\pi}{8} \right).