Logo
Jeeify
MCQMediumJEE 2025Modulus & Argument

JEE Mathematics 2025 Question with Solution

The number of complex numbers zz, satisfying ∣z∣=1|z| = 1 and ∣zz‾+z‾z∣=1,\left| \frac{z}{\overline{z}} + \frac{\overline{z}}{z} \right| = 1, is:

  • A

    66

  • B

    44

  • C

    1010

  • D

    88

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: ∣z∣=1|z| = 1 and ∣zz‾+z‾z∣=1\left| \frac{z}{\overline{z}} + \frac{\overline{z}}{z} \right| = 1

Find: The number of complex numbers zz satisfying both conditions.

Since ∣z∣=1|z| = 1, let

z=eiθ,z‾=e−iθz = e^{i\theta}, \quad \overline{z} = e^{-i\theta}

Then

zz‾=e2iθ,z‾z=e−2iθ\frac{z}{\overline{z}} = e^{2i\theta}, \quad \frac{\overline{z}}{z} = e^{-2i\theta}

So,

zz‾+z‾z=e2iθ+e−2iθ=2cos⁡(2θ)\frac{z}{\overline{z}} + \frac{\overline{z}}{z} = e^{2i\theta} + e^{-2i\theta} = 2\cos(2\theta)

Therefore the given condition becomes

∣2cos⁡(2θ)∣=1|2\cos(2\theta)| = 1

which gives

∣cos⁡(2θ)∣=12|\cos(2\theta)| = \frac{1}{2}

Hence,

cos⁡(2θ)=±12\cos(2\theta) = \pm \frac{1}{2}

The corresponding values are

2θ=π3,2π3,4π3,5π3,7π3,8π3,10π3,11π32\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}, \frac{11\pi}{3}

So,

θ=π6,π3,2π3,5π6,7π6,4π3,5π3,11π6\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{11\pi}{6}

Thus there are 88 distinct complex numbers on the unit circle satisfying the condition.

Therefore, the correct option is D.

Using trigonometric form

Given: ∣z∣=1|z| = 1.

Find: How many distinct values of zz satisfy ∣zz‾+z‾z∣=1\left| \frac{z}{\overline{z}} + \frac{\overline{z}}{z} \right| = 1.

Any complex number on the unit circle can be written as

z=a+ibwitha2+b2=1z = a + ib \quad \text{with} \quad a^2 + b^2 = 1

or equivalently in polar form as

z=eiθz = e^{i\theta}

Then its conjugate is

z‾=e−iθ\overline{z} = e^{-i\theta}

Now compute the two fractions:

zz‾=eiθe−iθ=e2iθ\frac{z}{\overline{z}} = \frac{e^{i\theta}}{e^{-i\theta}} = e^{2i\theta} z‾z=e−iθeiθ=e−2iθ\frac{\overline{z}}{z} = \frac{e^{-i\theta}}{e^{i\theta}} = e^{-2i\theta}

Adding,

e2iθ+e−2iθ=2cos⁡(2θ)e^{2i\theta} + e^{-2i\theta} = 2\cos(2\theta)

Hence the equation becomes

∣e2iθ+e−2iθ∣=1\left| e^{2i\theta} + e^{-2i\theta} \right| = 1

that is,

∣2cos⁡(2θ)∣=1|2\cos(2\theta)| = 1

So,

∣cos⁡(2θ)∣=12|\cos(2\theta)| = \frac{1}{2}

This means

cos⁡(2θ)=±12\cos(2\theta) = \pm \frac{1}{2}

There are 44 values of 2θ2\theta in one interval of length 2π2\pi. Since θ\theta runs over [0,2π)[0, 2\pi), the variable 2θ2\theta runs over an interval of length 4π4\pi, giving 88 distinct solutions.

Therefore, the number of complex numbers is 88, so the correct option is D.](streamdown:incomplete-link)

Common mistakes

  • Taking zz‾+z‾z\frac{z}{\overline{z}} + \frac{\overline{z}}{z} to be constant because ∣z∣=1|z| = 1. This is wrong because ∣z∣=1|z| = 1 only places zz on the unit circle; the argument θ\theta still varies. Write z=eiθz = e^{i\theta} and reduce the expression to 2cos⁡(2θ)2\cos(2\theta).

  • Counting only the solutions of cos⁡(2θ)=12\cos(2\theta) = \frac{1}{2} and forgetting cos⁡(2θ)=−12\cos(2\theta) = -\frac{1}{2}. This misses half the valid angles. Use ∣cos⁡(2θ)∣=12|\cos(2\theta)| = \frac{1}{2} carefully, which means both signs are allowed.

  • Finding the valid values of 2θ2\theta correctly but forgetting to divide by 22 to obtain θ\theta. This gives the wrong set of points on the unit circle. After solving for 2θ2\theta, convert each value to the corresponding θ\theta before counting distinct zz.

Practice more Modulus & Argument questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Modulus & Argument

Related questions