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MCQMediumJEE 2024Modulus & Argument

JEE Mathematics 2024 Question with Solution

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⁡(3π/8),3π/8)(2 \sec(3\pi/8), 3\pi/8)

  • B

    (2sec⁡(3π/8),5π/8)(2 \sec(3\pi/8), 5\pi/8)

  • C

    (2sec⁡(5π/8),3π/8)(2 \sec(5\pi/8), 3\pi/8)

  • D

    (2sec⁡(11π/8),11π/8)(2 \sec(11\pi/8), 11\pi/8)

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: z=2−i(2tan⁡5π8)z = 2 - i\left(2 \tan \frac{5\pi}{8}\right)

Find: modulus rr and amplitude θ\theta of zz.

Write

z=a+ibz = a + ib

where

a=2,b=−2tan⁡5π8.a = 2, \qquad b = -2 \tan \frac{5\pi}{8}.

Now the modulus is

r=a2+b2r = \sqrt{a^2 + b^2}

So,

r=22+(−2tan⁡5π8)2r = \sqrt{2^2 + \left(-2 \tan \frac{5\pi}{8}\right)^2} =4+4tan⁡25π8= \sqrt{4 + 4 \tan^2 \frac{5\pi}{8}} =21+tan⁡25π8= 2\sqrt{1 + \tan^2 \frac{5\pi}{8}}

Using 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x,

r=2sec⁡5π8r = 2 \sec \frac{5\pi}{8}

Also,

sec⁡(π−3π8)=sec⁡3π8\sec\left(\pi - \frac{3\pi}{8}\right) = \sec \frac{3\pi}{8}

Hence,

r=2sec⁡3π8.r = 2 \sec \frac{3\pi}{8}.

For the amplitude,

θ=tan⁡−1(ba)\theta = \tan^{-1}\left(\frac{b}{a}\right) =tan⁡−1(−2tan⁡5π82)= \tan^{-1}\left(\frac{-2 \tan \frac{5\pi}{8}}{2}\right) =tan⁡−1(−tan⁡5π8)= \tan^{-1}\left(-\tan \frac{5\pi}{8}\right)

Since the complex number lies in the second quadrant,

θ=π−5π8=3π8.\theta = \pi - \frac{5\pi}{8} = \frac{3\pi}{8}.

Therefore,

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

So the correct option is A.

Using coordinates of the complex number

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

Find: the pair (r,θ)(r, \theta).

Here,

x=2,y=−2tan⁡5π8.x = 2, \qquad y = -2 \tan \frac{5\pi}{8}.

Step 1: Calculate the modulus.

r=x2+y2r = \sqrt{x^2 + y^2} =(2)2+(2tan⁡5π8)2= \sqrt{(2)^2 + \left(2 \tan \frac{5\pi}{8}\right)^2} =2sec⁡5π8= 2 \sec \frac{5\pi}{8} =2sec⁡(π−3π8)= 2 \sec \left(\pi - \frac{3\pi}{8}\right) =2sec⁡3π8.= 2 \sec \frac{3\pi}{8}.

Step 2: Calculate the amplitude.

θ=tan⁡−1(−2tan⁡5π82)\theta = \tan^{-1}\left(\frac{-2 \tan \frac{5\pi}{8}}{2}\right) =tan⁡−1(−tan⁡5π8)= \tan^{-1}\left(-\tan \frac{5\pi}{8}\right) =3π8.= \frac{3\pi}{8}.

Therefore,

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

and the correct answer is A.

Common mistakes

  • Using θ=tan⁡−1(b/a)\theta = \tan^{-1}(b/a) without checking the quadrant. Here a>0a > 0 and b<0b < 0, so the sign of the imaginary part must be handled carefully. Always identify the quadrant before fixing the amplitude.

  • Leaving the modulus as 2sec⁡5π82 \sec \frac{5\pi}{8} without converting it to the positive principal form used in the options. The modulus must be positive, so rewrite it consistently as 2sec⁡3π82 \sec \frac{3\pi}{8}.

  • Treating the coefficient of ii incorrectly. In z=2−i(2tan⁡5π8)z = 2 - i\left(2 \tan \frac{5\pi}{8}\right), the imaginary part is −2tan⁡5π8-2 \tan \frac{5\pi}{8}, not +2tan⁡5π8+2 \tan \frac{5\pi}{8}. First rewrite the number in the form a+iba + ib.

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