Let a complex number be $$w = 1 - \sqrt{3}i$$. Let another complex number $$z$$ be such that $$|zw| = 1$$ and $$\arg(z) - \arg(w) = \frac{\pi}{2}$$. Then the area of the triangle (in sq. units) with vertices origin, $$z$$ and $$w$$ is equal to

We have $$w = 1 - \sqrt{3}\,i$$. First, $$|w| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$, and $$\arg(w) = -\frac{\pi}{3}$$ (since $$w$$ lies in the fourth quadrant with $$\tan^{-1}\frac{\sqrt{3}}{1} = \frac{\pi}{3}$$).

We need $$|zw| = 1$$, so $$|z| \cdot |w| = 1$$, giving $$|z| = \frac{1}{|w|} = \frac{1}{2}$$. Also $$\arg(z) - \arg(w) = \frac{\pi}{2}$$, so $$\arg(z) = \arg(w) + \frac{\pi}{2} = -\frac{\pi}{3} + \frac{\pi}{2} = \frac{\pi}{6}$$.

Therefore $$z = \frac{1}{2}\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = \frac{1}{2}\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = \frac{\sqrt{3}}{4} + \frac{i}{4}$$.

The area of the triangle with vertices at the origin $$O = (0,0)$$, $$z = \left(\frac{\sqrt{3}}{4}, \frac{1}{4}\right)$$, and $$w = (1, -\sqrt{3})$$ is given by $$\text{Area} = \frac{1}{2}|x_z \cdot y_w - x_w \cdot y_z|$$. Substituting: $$\text{Area} = \frac{1}{2}\left|\frac{\sqrt{3}}{4} \cdot (-\sqrt{3}) - 1 \cdot \frac{1}{4}\right| = \frac{1}{2}\left|-\frac{3}{4} - \frac{1}{4}\right| = \frac{1}{2} \times 1 = \frac{1}{2}$$.

The area of the triangle is $$\frac{1}{2}$$ square units, which is option (2).