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.

Answer 2

1. Since $|z|=1$, we can write each complex number in exponential form: $$ z_k = e^{i\theta_k},\ \text{with}\ \theta_1=-\tfrac{\pi}{4},\ \theta_2=0,\ \theta_3=\tfrac{\pi}{4}. $$ 
2. Using $\overline{z_k}=e^{-i\theta_k}$, compute each term:
   • $z_1\overline{z_2} = e^{i(\theta_1 - \theta_2)} = e^{-i\pi/4}$
   • $z_2\overline{z_3} = e^{i(\theta_2 - \theta_3)} = e^{-i\pi/4}$
   • $z_3\overline{z_1} = e^{i(\theta_3 - \theta_1)} = e^{i(\pi/2)} = i$
3. Therefore, $$ z_1\overline{z_2} + z_2\overline{z_3} + z_3\overline{z_1} = e^{-i\pi/4} + e^{-i\pi/4} + i = 2e^{-i\pi/4} + i. $$
4. Now express $e^{-i\pi/4}$ in rectangular form: $e^{-i\pi/4} = \cos(-\pi/4) + i\sin(-\pi/4) = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$.
5. Substitute: $$ 2e^{-i\pi/4} + i = 2\left(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) + i = \sqrt{2} - i\sqrt{2} + i = \sqrt{2} + i(1 - \sqrt{2}). $$
6. Compute the modulus squared: $$ |z_1\overline{z_2} + z_2\overline{z_3} + z_3\overline{z_1}|^2 = (\sqrt{2})^2 + (1 - \sqrt{2})^2 = 2 + (1 - 2\sqrt{2} + 2) = 5 - 2\sqrt{2}. $$
7. It is given that this equals $\alpha + \beta^2$, where $\alpha,\beta \in \mathbb{Z}$. 
   Comparing with $5 - 2\sqrt{2}$, we can identify: $$\alpha = 5,\quad \beta = -1.$$
8. Hence, $$ \alpha^2 + \beta^2 = 5^2 + (-1)^2 = 25 + 1 = 26. $$ But $26$ is not among the options, so let's check again.
9. Wait — observe carefully: The expression was $|A|^2 = \alpha + \beta^2$, so $\alpha$ and $\beta$ must be integers such that $5 - 2\sqrt{2} = \alpha + \beta^2$ is impossible (since $\sqrt{2}$ is irrational). That suggests we made a misinterpretation — let’s re-evaluate algebraically.
10. Let's compute the exact real and imaginary parts and square the modulus directly without introducing irrational separation. 
    $$ A = 2e^{-i\pi/4} + i = 2(\cos(-\pi/4) + i\sin(-\pi/4)) + i = 2\left(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) + i = \sqrt{2} + i(1 - \sqrt{2}). $$ Therefore, $|A|^2 = (\sqrt{2})^2 + (1 - \sqrt{2})^2 = 2 + 1 - 2\sqrt{2} + 2 = 5 - 2\sqrt{2}$. (Confirmed correct.)
11. Hence we can express numerically: $5 - 2\sqrt{2} \approx 5 - 2(1.414) = 5 - 2.828 = 2.172$. Let's check which option combination of $\alpha,\beta$ fits this numeric value:
    • If $\alpha=1,\ \beta=1$, $\alpha+\beta^2=2$ ✅ close to 2.17.
12. So $\alpha^2+\beta^2 = 1^2 + 1^2 = 2$ — not among the options. Let's double-check if we perhaps misread the given expression; maybe it was $|A|^2 = \alpha + \beta \sqrt{2}$ instead of $\alpha + \beta^2$. (Given pattern, it's likely meant to be $\alpha + \beta\sqrt{2}$ with integers $\alpha,\beta$.)
13. If $|A|^2 = \alpha + \beta\sqrt{2}$, then comparing with $5 - 2\sqrt{2}$ gives $\alpha = 5,\ \beta = -2$. Then $$ \alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29. $$

Answer

$\alpha^2 + \beta^2 = 29.$ (Option 4)