Question:

<p><strong>Step 1:</strong> From $\bar{z} + i\bar{\omega} = 0$, we get $\bar{z} = -i\bar{\omega}$</p><p><strong>Step 2:</strong> Taking conjugates: $z = -i\omega$, so $\omega = \frac{z}{-i} = \frac{z \cdot i}{-i \cdot i} = \frac{iz}{1} = iz$</p><p><strong>Step 3:</strong> Therefore $z\omega = z \cdot iz = iz^2$</p><p><strong>Step 4:</strong> Given $\arg(z\omega) = \pi$, we have $\arg(iz^2) = \pi$</p><p><strong>Step 5:</strong> Since $\arg(iz^2) = \arg(i) + \arg(z^2) = \frac{\pi}{2} + 2\arg(z) = \pi$ (taking principal value)</p><p><strong>Step 6:</strong> This gives $2\arg(z) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$</p><p><strong>Step 7:</strong> Therefore $\arg(z) = \frac{\pi}{4}$</p><p>∴ Answer: C</p>