Let $z = 1 + \sqrt{3}\mathrm{i}$, $\arg(w) = \frac{\pi}{4}$ and $|zw| = 2\sqrt{2}$.

1. Calculate the modulus of $z$.

$|z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$

2. Determine the modulus of $w$ using the multiplicative property of moduli.

$|zw| = |z||w|$
$2|w| = 2\sqrt{2} \implies |w| = \sqrt{2}$

3. Convert $w$ from modulus-argument form to Cartesian form.

$w = |w|\left(\cos(\arg(w)) + \mathrm{i}\sin(\arg(w))\right)$
$w = \sqrt{2}\left(\cos\frac{\pi}{4} + \mathrm{i}\sin\frac{\pi}{4}\right)$
$w = \sqrt{2}\left(\frac{1}{\sqrt{2}} + \mathrm{i}\frac{1}{\sqrt{2}}\right) = 1 + \mathrm{i}$

1. Find the argument of $zw$.

$\arg(z) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$
$\arg(zw) = \arg(z) + \arg(w)$
$\arg(zw) = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$

2. Set up the condition for $(zw)^n$ to be purely imaginary.

For $(zw)^n$ to be purely imaginary, its argument must be an odd multiple of $\frac{\pi}{2}$.
$\arg\left((zw)^n\right) = n \arg(zw) = \frac{7n\pi}{12}$
$\frac{7n\pi}{12} = \frac{\pi}{2} + k\pi$, for $k \in \mathbb{Z}$

3. Solve for the least positive integer $n$.

$\frac{7n}{12} = \frac{2k+1}{2}$
$7n = 6(2k+1)$
For $n$ to be a positive integer, $2k+1$ must be a multiple of $7$. The smallest positive integer value for $2k+1$ that is a multiple of $7$ is $7$ (when $k=3$).
$7n = 6(7) \implies n = 6$