<p>If \(z_1 = a + ib\) and \(z_2 = c + id\) are complex numbers such that \(|z_1| = |z_2| = 1\) and \(\text{Re}(z_1\overline{z_2}) = 0\), where \(i = \sqrt{-1}\), then the complex numbers \(w_1 = a + ic\) and \(w_2 = b + id\) satisfy</p>
Step-by-Step Solution
Key Concept: When $|z_1| = |z_2| = 1$, we can write them in exponential form. The orthogonality condition constrains their arguments and creates a structured relationship between the components.
<p><strong>Given:</strong> $|z_1| = 1$, $|z_2| = 1$, and $\text{Re}(z_1\overline{z_2}) = 0$</p><p><strong>Step 1:</strong> Since $|z_1| = 1$, we have $z_1 = e^{i\alpha}$ for some angle $\alpha$. Thus $a = \cos\alpha$ and $b = \sin\alpha$.</p><p><strong>Step 2:</strong> Since $|z_2| = 1$, we have $z_2 = e^{i\beta}$ for some angle $\beta$. Thus $c = \cos\beta$ and $d = \sin\beta$.</p><p><strong>Step 3:</strong> The condition $\text{Re}(z_1\overline{z_2}) = 0$ gives $\text{Re}(e^{i(\alpha-\beta)}) = \cos(\alpha - \beta) = 0$, so $\alpha - \beta = \pm\frac{\pi}{2}$.</p><p><strong>Step 4:</strong> Let $\beta = \alpha - \frac{\pi}{2}$. Then $c = \sin\alpha$ and $d = -\cos\alpha$.</p><p><strong>Step 5:</strong> Now $w_1 = a + ic = \cos\alpha + i\sin\alpha = e^{i\alpha}$, so $|w_1| = 1$ ✓</p><p><strong>Step 6:</strong> And $w_2 = b + id = \sin\alpha - i\cos\alpha = -ie^{i\alpha}$, so $|w_2| = 1$ ✓</p><p><strong>Step 7:</strong> $w_1\overline{w_2} = e^{i\alpha} \cdot ie^{-i\alpha} = i$, which is purely imaginary, so $\text{Re}(w_1\overline{w_2}) = 0$ ✓</p><p>∴ Options (a), (b), and (c) are correct.</p>
Correct Answer: a, b, c