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JEE Main 202429 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2024 · 29 January · Evening Shift · Mathematics · Complex Number

Let r and θ respectively be the modulus and amplitude of the...

Let rr and θ\theta respectively be the modulus and amplitude of the complex number z=2−i(2tan5π8)z = 2 - i\left(2tan\dfrac{5\pi}{8}\right), then (r,θ)( r, \theta ) is equal to

Options

  1. A(2sec3π8,3π8)\left(2sec\dfrac{3\pi}{8}, \dfrac{3\pi}{8}\right)
  2. B(2sec3π8,5π8)\left(2sec\dfrac{3\pi}{8}, \dfrac{5\pi}{8}\right)
  3. C(2sec5π8,3π8)\left(2sec\dfrac{5\pi}{8}, \dfrac{3\pi}{8}\right)
  4. D(2sec11π8,11π8)\left(2sec\dfrac{11\pi}{8}, \dfrac{11\pi}{8}\right)

Correct answer

A. (2sec3π8,3π8)\left(2sec\dfrac{3\pi}{8}, \dfrac{3\pi}{8}\right)

Step-by-step solution

Given: z=2−i(2tan5π8)z = 2 - i\left(2tan\dfrac{5\pi}{8}\right)

⇒z=2[1−itan(5π8)]⇒ z = 2\left[1 - itan\left(\dfrac{5\pi}{8}\right)\right]

⇒z=2[1−isin(5π8)cos⁡(5π8)]⇒ z = 2\left[1 - \dfrac{isin\left(\dfrac{5\pi}{8}\right)}{\cos \left(\dfrac{5\pi}{8}\right)}\right]

⇒z=2cos⁡(5π8)[cos⁡(5π8)−isin(5π8)]⇒ z = \dfrac{2}{\cos \left(\dfrac{5\pi}{8}\right)}\left[\cos \left(\dfrac{5\pi}{8}\right) - isin\left(\dfrac{5\pi}{8}\right)\right]

⇒z=2cos⁡(π−3π8)[cos⁡(π−3π8)−isin(π−3π8)]⇒ z = \dfrac{2}{\cos \left(\pi - \dfrac{3\pi}{8}\right)}\left[\cos \left(\pi - \dfrac{3\pi}{8}\right) - isin\left(\pi - \dfrac{3\pi}{8}\right)\right]

⇒z=2−cos⁡(3π8)[−cos⁡(3π8)−isin(3π8)]⇒ z = \dfrac{2}{- \cos \left(\dfrac{3\pi}{8}\right)}\left[- \cos \left(\dfrac{3\pi}{8}\right) - isin\left(\dfrac{3\pi}{8}\right)\right]

⇒z=2cos⁡(3π8)[cos⁡(3π8)+isin(3π8)]⇒ z = \dfrac{2}{\cos \left(\dfrac{3\pi}{8}\right)}\left[\cos \left(\dfrac{3\pi}{8}\right) + isin\left(\dfrac{3\pi}{8}\right)\right]

⇒z=2sec3π8ei⋅3π8⇒ z = 2sec\dfrac{3\pi}{8}{e}^{i \cdot \dfrac{3\pi}{8}}

Now on comparing with z=∣z∣eiθz = \left|z\right|{e}^{i\theta} we get,

⇒θ=3π8,r=2sec3π8⇒ \theta = \dfrac{3\pi}{8}, r = 2sec\dfrac{3\pi}{8}

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