Question

Let \( z_1, z_2, z_3 \) be three complex numbers on the circle \( |z| = 1 \) with \( \arg(z_1) = -\frac{\pi}{4}, \arg(z_2) = 0 \) and \( \arg(z_3) = \frac{\pi}{4} \). If \( |z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 = \alpha + \beta \sqrt{2} \), where \( \alpha, \beta \in \mathbb{Z} \), then the value of \( \alpha^2 + \beta^2 \) is :

• A. 24
• B. 41
• C. 31
• D. 29

Hint:

For expressions involving complex numbers, use polar form and properties of modulus to simplify.

Answer 1

The Correct Option is D

Given 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 calculate the expression \(|z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 = \alpha + \beta \sqrt{2}\).
1. Express \(z_i\) as \(z_1 = e^{-i\frac{\pi}{4}} = \frac{1}{\sqrt{2}}(1-i)\), \(z_2 = e^{0} = 1\), \(z_3 = e^{i\frac{\pi}{4}} = \frac{1}{\sqrt{2}}(1+i)\).
2. Conjugates: \(\overline{z_2} = 1\), \(\overline{z_3} = \frac{1}{\sqrt{2}}(1-i)\), \(\overline{z_1} = \frac{1}{\sqrt{2}}(1+i)\).
3. Compute the expression:
\[\begin{align*} z_1 \overline{z_2} &= z_1 = \frac{1}{\sqrt{2}}(1-i), \\ z_2 \overline{z_3} &= \frac{1}{\sqrt{2}}(1-i), \\ z_3 \overline{z_1} &= \frac{1}{\sqrt{2}}(1+i)\cdot\frac{1}{\sqrt{2}}(1-i) = 1. \end{align*}\]
4. Summing these:\[ z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1} = \frac{1}{\sqrt{2}}(1-i) + \frac{1}{\sqrt{2}}(1-i) + 1 = \sqrt{2}(1-i) + 1 = (\sqrt{2} + 1) - i\sqrt{2}. \]
5. Find the magnitude squared:\[ |(\sqrt{2} + 1) - i\sqrt{2}|^2 = ((\sqrt{2} + 1)^2 + (\sqrt{2})^2) = 3 + 2\sqrt{2} + 2 = 5 + 2\sqrt{2}. \]
6. Thus, \(\alpha = 5\) and \(\beta = 2\), so \(\alpha^2 + \beta^2 = 5^2 + 2^2 = 25 + 4 = 29\).
Therefore, the value of \(\alpha^2 + \beta^2\) is 29.