QJEEM25Ma014f: JEE Main 2025 Mathematics - Class 11 JEEM25Ma014f Solution: Let and be three complex numbers on the circle with and . If ,…

To solve the problem, we start with the given information about the complex numbers $z_1, z_2,$ and $z_3$ , which lie on the unit circle $|z| = 1$ . Their arguments are as follows:

$\arg(z_1) = -\frac{\pi}{4}$

$\arg(z_2) = 0$

$\arg(z_3) = \frac{\pi}{4}$

Thus, the complex numbers can be represented as:

$z_1 = e^{-i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$

$z_2 = e^{i\cdot 0} = 1$

$z_3 = e^{i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$

Next, calculate the conjugates needed:

$\bar{z}_2 = 1$

$\bar{z}_3 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$

$\bar{z}_1 = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$

We need to evaluate:

$z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1$

Calculate each term separately:

$z_1 \bar{z}_2 = \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) \cdot 1 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$

$z_2 \bar{z}_3 = 1 \cdot \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$

$z_3 \bar{z}_1 = \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) \cdot \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = \frac{(1 + i)^2}{2} = \frac{1 + 2i - 1}{2} = i$

Sum the evaluated terms:

$z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 = \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) + i$

Simplify:

$= \sqrt{2} + i - \sqrt{2}i = \sqrt{2} + i(1-\sqrt{2})$

Calculate the modulus squared:

$\left| \sqrt{2} + i (1 - \sqrt{2}) \right|^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2$

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

Thus, the expression $|z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1|^2$ simplifies as follows:

$\alpha = 5$

$\beta = -2$

Finally, compute $\alpha^2 + \beta^2$ :

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

Therefore, the value of $\alpha^2 + \beta^2$ is 29.