1. Write $w$ in polar form

Given $w=1-\sqrt{3}\,i$

Its modulus is $|w|=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{1+3}=2$

Its argument is in the fourth quadrant, and $\tan \theta=\frac{-\sqrt{3}}{1}=-\sqrt{3}$ so $\arg(w)=-\frac{\pi}{3}$

Hence, $w=2\left(\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right)$

2. Use the condition $|zw|=1$

We know $|zw|=|z|\,|w|=1$ Since $|w|=2$, we get $|z|\cdot 2=1\implies |z|=\frac{1}{2}$

3. Use the argument condition

Given $\arg(z)-\arg(w)=\frac{\pi}{2}$ Therefore, $\arg(z)=\arg(w)+\frac{\pi}{2}=-\frac{\pi}{3}+\frac{\pi}{2}=\frac{\pi}{6}$

So $z$ has modulus $\frac12$ and argument $\frac{\pi}{6}$.

4. Find the angle between $z$ and $w$

The vectors from the origin to $z$ and $w$ make an angle $\left|\arg(z)-\arg(w)\right|=\frac{\pi}{2}$

So the triangle formed by the origin, $z$, and $w$ has two sides of lengths $|z|=\frac12,\qquad |w|=2$ and included angle $\frac{\pi}{2}$

5. Compute the area

Area of triangle with sides $|z|,|w|$ and included angle $\theta$ is $\text{Area}=\frac12 |z||w|\sin\theta$

Thus, $\text{Area}=\frac12\cdot \frac12\cdot 2\cdot \sin\frac{\pi}{2}$ $=\frac12\cdot 1\cdot 1=\frac12$

6. Match with options

The area is $\boxed{\frac12}$ So the correct option is D.