Given three complex numbers \(z_1, z_2, z_3\) on the unit circle \(|z| = 1\) with arguments \(\arg(z_1) = -\frac{\pi}{4}\), \(\arg(z_2) = 0\), and \(\arg(z_3) = \frac{\pi}{4}\), we evaluate \(|z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 = \alpha + \beta \sqrt{2}\).
1. Express \(z_i\) in polar and Cartesian forms: \(z_1 = e^{-i\frac{\pi}{4}} = \frac{1}{\sqrt{2}}(1-i)\), \(z_2 = e^{0} = 1\), \(z_3 = e^{i\frac{\pi}{4}} = \frac{1}{\sqrt{2}}(1+i)\).
2. Determine the conjugates: \(\overline{z_2} = 1\), \(\overline{z_3} = \frac{1}{\sqrt{2}}(1-i)\), \(\overline{z_1} = \frac{1}{\sqrt{2}}(1+i)\).
3. Calculate the products:
\[\begin{align*} z_1 \overline{z_2} &= \frac{1}{\sqrt{2}}(1-i), \\ z_2 \overline{z_3} &= \frac{1}{\sqrt{2}}(1-i), \\ z_3 \overline{z_1} &= \left(\frac{1}{\sqrt{2}}(1+i)\right) \left(\frac{1}{\sqrt{2}}(1+i)\right) = \frac{1}{2}(1+2i-1) = i \cdot 1 = i \end{align*}\]
4. Sum the products:\[ z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1} = \frac{1}{\sqrt{2}}(1-i) + \frac{1}{\sqrt{2}}(1-i) + i = \frac{2}{\sqrt{2}}(1-i) + i = \sqrt{2}(1-i) + i = \sqrt{2} - i\sqrt{2} + i = \sqrt{2} + i(1-\sqrt{2}). \]
5. Compute the magnitude squared:\[ |\sqrt{2} + i(1-\sqrt{2})|^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2 = 2 + (1 - 2\sqrt{2} + 2) = 2 + 3 - 2\sqrt{2} = 5 - 2\sqrt{2}. \]
6. Comparing \(5 - 2\sqrt{2}\) with \(\alpha + \beta \sqrt{2}\), we identify \(\alpha = 5\) and \(\beta = -2\). The problem asks for \(\alpha^2 + \beta^2\). Thus, \(\alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29\).
The value of \(\alpha^2 + \beta^2\) is 29.