Complex Numbers
Squared Modulus of Sum of Conjugate Products
nta_pyq_2025_apr
Grade 11
Question:
Let $z_1$, $z_2$, $z_3$ be three complex numbers lying on the circle $|z|=1$ with $\arg(z_1)=-\pi/4$, $\arg(z_2)=0$, $\arg(z_3)=\pi/4$. If $|z_1\bar{z}_2+z_2\bar{z}_3+z_3\bar{z}_1|^2 = \alpha+\beta\sqrt{2}$ where $\alpha,\beta\in\mathbb{Z}$, then $\alpha^2+\beta^2$ equals
Step-by-Step Solution
Key Concept: Write $z_k=e^{i\theta_k}$ so that $z_j\bar{z}_k=e^{i(\theta_j-\theta_k)}$; sum the three exponentials, then compute $|\cdot|^2 = w\bar{w}$.
$z_1=e^{-i\pi/4}$, $z_2=1$, $z_3=e^{i\pi/4}$.
$z_1\bar{z}_2 = e^{-i\pi/4} = \dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}i$.
$z_2\bar{z}_3 = e^{-i\pi/4} = \dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}i$.
$z_3\bar{z}_1 = e^{i\pi/2} = i$.
Sum $= \sqrt{2} + (1-\sqrt{2})i$.
$|\text{Sum}|^2 = (\sqrt{2})^2+(1-\sqrt{2})^2 = 2+1-2\sqrt{2}+2 = 5-2\sqrt{2}$.
So $\alpha=5$, $\beta=-2$, and $\alpha^2+\beta^2=25+4=29$.
Correct Answer: 2