Two of the roots of the equation $z^{4}-kz^{2}+k^{3}z-k^{4}=0$, where $k$ is a positive real number, are $z_{1}=k$ and $z_{2}=-k$. The other two roots are denoted $z_{3}$ and $z_{4}$, where $\mathrm{Im}(z_{3})>0$ and $\mathrm{Im}(z_{4})<0$.

1. Calculate the modulus of $z_3$.

$|z_3| = \sqrt{\left(\frac{k}{2}\right)^2 + \left(\frac{\sqrt{3}k}{2}\right)^2}$
$|z_3| = \sqrt{\frac{k^2}{4} + \frac{3k^2}{4}} = \sqrt{k^2}$
Since $k$ is a positive real number, $|z_3| = k$.

2. Calculate the argument of $z_3$.

Since both the real and imaginary parts of $z_3$ are positive, it lies in the first quadrant in the Argand diagram.
$\arg z_3 = \tan^{-1}\left(\frac{\frac{\sqrt{3}k}{2}}{\frac{k}{2}}\right)$
$\arg z_3 = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$

1. Find the argument of $\overline{z_3}$ and $-1+i$.

From part (i), $z_3 = \frac{k}{2} + \frac{\sqrt{3}k}{2}i = k\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$.
The conjugate is $\overline{z_3} = k\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)$, so $\arg(\overline{z_3}) = -\frac{\pi}{3}$.
For the denominator $-1+i$, its argument is $\pi - \tan^{-1}(1) = \frac{3\pi}{4}$.
Thus, $\arg\left(\frac{\overline{z_3}}{-1+i}\right) = \arg(\overline{z_3}) - \arg(-1+i) = -\frac{\pi}{3} - \frac{3\pi}{4} = -\frac{13\pi}{12}$.
This is equivalent to a principal argument of $\frac{11\pi}{12}$.

2. Formulate the condition for the expression to be purely imaginary.

Let $w = \frac{\overline{z_3}}{-1+i}$. For $w^n$ to be purely imaginary, its real part must be zero, which means its argument must be an odd multiple of $\frac{\pi}{2}$.
$\arg(w^n) = n\arg(w) = \frac{11n\pi}{12}$.
We require $\frac{11n\pi}{12} = \frac{\pi}{2} + m\pi = \frac{(2m+1)\pi}{2}$ for some integer $m$.

3. Solve for the two smallest positive integer values of $n$.

Simplifying the equation gives:
$11n = 6(2m+1)$
This requires $11n$ to be an odd multiple of $6$. Since $\gcd(11, 6) = 1$, $n$ must be a multiple of $6$.
Let $n = 6p$, where $p$ is an integer. Substituting this gives:
$11(6p) = 6(2m+1) \implies 11p = 2m+1$
Since $2m+1$ represents an odd integer, $11p$ must be odd, which means $p$ must be an odd integer.
For $n$ to be the two smallest positive integers, we choose the two smallest positive odd integers for $p$: $p = 1$ and $p = 3$.
For $p = 1$: $n = 6(1) = 6$
For $p = 3$: $n = 6(3) = 18$