JEE Main 2024 · 29 January · evening · Single correct · +4 / −1

Let r and θ respectively be the modulus and amplitude of the complex number z=2-i(2 tan (5 π)/8), then (r, θ) is equal to

  1. (a) (2 sec (11 π)/8, (11 π)/8)
  2. (b) (2 sec (3 π)/8, (3 π)/8)
  3. (c) (2 sec (5 π)/8, (3 π)/8)
  4. (d) (2 sec (3 π)/8, (5 π)/8)
Answer and solution

Answer (b) (2 sec (3 π)/8, (3 π)/8)

Solution

Maths sir · 4 steps

First turn tan 5π/8 into -tan 3π/8; then the number sits in the first quadrant.

  1. Fix the sign. Since 5π/8 = π - 3π/8, we get tan 5π/8 = -tan 3π/8. So z = 2 + 2i tan 3π/8, and tan 3π/8 > 0 because 3π/8 is an acute angle.
  2. Polar form. Take out 2 and write tan = sin/cos. This puts z in the form r(cos θ + i sin θ).

    z = 2(1 + i tan 3π/8) = 2 sec 3π/8(cos 3π/8 + i sin 3π/8)

  3. Check positivity. cos 3π/8 > 0, so 2 sec 3π/8 is positive and is a valid modulus. The real and imaginary parts are both positive, so the argument 3π/8 lies in the first quadrant, as it should.
  4. So. r = 2 sec 3π/8 and θ = 3π/8. That is option (b).

The idea to keep. Before you read off r and θ, make sure the factor in front is positive and the angle matches the quadrant of the point. A modulus can never be negative.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Complex Numbers

Asked about 160 times in JEE Main since 2020. This question is from 2024.