Let $z_1, z_2$ and $z_3$ 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 $|z_1\bar{z}_2 + z_2\bar{z}_3 + z_3\bar{z}_1|^2 = \alpha + \beta\sqrt{2}$, where $\alpha, \beta \in \mathbb{Z}$, then the value of $\alpha^2 + \beta^2$ is:
- (1) 24
- (2) 41
- (3) 31
- (4) 29
Solution:
Given $|z| = 1 \Rightarrow z\bar{z} = 1 \Rightarrow \bar{z} = \frac{1}{z}$. Also $z = e^{i\theta}$.
From the given arguments:
$$ z_1 = e^{-i\frac{\pi}{4}}, \quad z_2 = e^{i(0)} = 1, \quad z_3 = e^{i\frac{\pi}{4}} $$
Now consider the expression inside the modulus:
$$ E = z_1\bar{z}_2 + z_2\bar{z}_3 + z_3\bar{z}_1 $$
$$ = e^{-i\frac{\pi}{4}} \cdot 1 + 1 \cdot e^{-i\frac{\pi}{4}} + e^{i\frac{\pi}{4}} \cdot e^{i\frac{\pi}{4}} $$
$$ = 2e^{-i\frac{\pi}{4}} + e^{i\frac{\pi}{2}} $$
$$ = 2\left(\cos\frac{\pi}{4} – i\sin\frac{\pi}{4}\right) + i $$
$$ = 2\left(\frac{1}{\sqrt{2}} – \frac{i}{\sqrt{2}}\right) + i = \sqrt{2} – i\sqrt{2} + i $$
$$ E = \sqrt{2} + i(1 – \sqrt{2}) $$
Now find $|E|^2$:
$$ |E|^2 = (\sqrt{2})^2 + (1 – \sqrt{2})^2 $$
$$ = 2 + (1 – 2\sqrt{2} + 2) = 5 – 2\sqrt{2} $$
Comparing with $\alpha + \beta\sqrt{2}$:
$$ \alpha = 5, \quad \beta = -2 $$
Required value $\alpha^2 + \beta^2 = (5)^2 + (-2)^2 = 25 + 4 = 29$.
Ans. (4)