JEE Mathematics 2025 Question with Solution

Given: $z_1, z_2, z_3$ lie on the unit circle with $\arg(z_1) = -\frac{\pi}{4}$, $\arg(z_2) = 0$, and $\arg(z_3) = \frac{\pi}{4}$.

Find: $\alpha^2 + \beta^2$ if

$$|z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 = \alpha + \beta \sqrt{2}$$

Write the complex numbers in polar form:

$$z_1 = e^{-i\pi/4} = \frac{1}{\sqrt{2}}(1-i), \qquad z_2 = 1, \qquad z_3 = e^{i\pi/4} = \frac{1}{\sqrt{2}}(1+i)$$

Their conjugates are:

$$\overline{z_2} = 1, \qquad \overline{z_3} = \frac{1}{\sqrt{2}}(1-i), \qquad \overline{z_1} = \frac{1}{\sqrt{2}}(1+i)$$

Now compute each term:

$$z_1\overline{z_2} = \frac{1}{\sqrt{2}}(1-i)$$ $$z_2\overline{z_3} = \frac{1}{\sqrt{2}}(1-i)$$ $$z_3\overline{z_1} = \frac{1}{\sqrt{2}}(1+i) \cdot \frac{1}{\sqrt{2}}(1+i) = i$$

So,

$$z_1\overline{z_2} + z_2\overline{z_3} + z_3\overline{z_1} = \frac{1}{\sqrt{2}}(1-i) + \frac{1}{\sqrt{2}}(1-i) + i = \sqrt{2} + i(1-\sqrt{2})$$

Hence,

$$\left|z_1\overline{z_2} + z_2\overline{z_3} + z_3\overline{z_1}\right|^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2 = 2 + 1 - 2\sqrt{2} + 2 = 5 - 2\sqrt{2}$$

Comparing with $\alpha + \beta\sqrt{2}$, we get

$$\alpha = 5, \qquad \beta = -2$$

Therefore,

$$\alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29$$

Therefore, the correct option is D.

Given: $z_k = e^{i\theta_k}$ with $\theta_1 = -\frac{\pi}{4}, \theta_2 = 0, \theta_3 = \frac{\pi}{4}$.

Find: $\alpha^2 + \beta^2$.

Since $\overline{z_k} = e^{-i\theta_k}$,

$$z_1\overline{z_2} = e^{i(\theta_1-\theta_2)} = e^{-i\pi/4}$$ $$z_2\overline{z_3} = e^{i(\theta_2-\theta_3)} = e^{-i\pi/4}$$ $$z_3\overline{z_1} = e^{i(\theta_3-\theta_1)} = e^{i\pi/2} = i$$

Thus,

$$A = z_1\overline{z_2} + z_2\overline{z_3} + z_3\overline{z_1} = 2e^{-i\pi/4} + i$$

Now,

$$e^{-i\pi/4} = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$$

So,

$$A = 2\left(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) + i = \sqrt{2} + i(1-\sqrt{2})$$

Therefore,

$$|A|^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2 = 5 - 2\sqrt{2}$$

Hence,

$$\alpha = 5, \qquad \beta = -2$$

and

$$\alpha^2 + \beta^2 = 29$$

So the correct option is D.

The alternative working shown on the page contains an intermediate inconsistency, but its final comparison with $\alpha + \beta\sqrt{2}$ also leads to the same answer 29.