The complex number $z$ is given by $x + i y$, where $x$ and $y$ are real numbers.

1. Expand $z^2$ in Cartesian form.

$z^2 = (x + iy)^2 \\ = x^2 + 2ixy + i^2y^2 \\ = (x^2 - y^2) + i(2xy)$

2. Use the given argument condition to determine the relationship between $x$ and $y$.

Since $\arg(z^2) = -\frac{\pi}{2}$, $z^2$ must lie on the negative imaginary axis. This means its real part is zero and its imaginary part is strictly negative.

$\text{Re}(z^2) = 0 \implies x^2 - y^2 = 0 \implies y^2 = x^2 \implies y = \pm x$

$\text{Im}(z^2) < 0 \implies 2xy < 0 \implies xy < 0$

For $xy$ to be negative, $x$ and $y$ must have opposite signs. Thus, we reject $y = x$ and take $y = -x$.

Additionally, since $z^2 \neq 0$ (as the argument is defined), $x \neq 0$.

1. Find the values of $x$ and $y$ using the implied result from part (i) and the given conditions.

From part (i), solving $\arg(z^2) = -\frac{\pi}{2}$ gives the relationship $y = -x$.
Given $|z| = 2$, we have $x^2 + y^2 = 2^2 = 4$.
Substituting $y = -x$ gives $x^2 + (-x)^2 = 4 \implies 2x^2 = 4 \implies x^2 = 2$.
Since $\text{Re}(z) > 0$, we know $x > 0$, so $x = \sqrt{2}$.
Then $y = -\sqrt{2}$.

2. Express $z$ and $z^*$ in exponential form to facilitate finding powers.

$z = \sqrt{2} - i\sqrt{2}$
We know $|z| = 2$ and $\arg(z) = \tan^{-1}\left(\frac{-\sqrt{2}}{\sqrt{2}}\right) = -\frac{\pi}{4}$.
Thus, $z = 2e^{-i\frac{\pi}{4}}$.
The complex conjugate is $z^* = 2e^{i\frac{\pi}{4}}$.

3. Simplify the expression $\frac{z^*}{z^n}$ using exponential form.

$\frac{z^*}{z^n} = \frac{2e^{i\frac{\pi}{4}}}{\left(2e^{-i\frac{\pi}{4}}\right)^n}$
$= \frac{2e^{i\frac{\pi}{4}}}{2^n e^{-in\frac{\pi}{4}}}$
$= 2^{1-n} e^{i\left(\frac{\pi}{4} - \left(-\frac{n\pi}{4}\right)\right)}$
$= 2^{1-n} e^{i\frac{(n+1)\pi}{4}}$

4. Find the smallest positive integer $n$ for which the expression is a negative real number.

For $\frac{z^*}{z^n}$ to be a negative real number, its argument must be an odd multiple of $\pi$.
$\frac{(n+1)\pi}{4} = (2k+1)\pi$ for $k \in \mathbb{Z}$.
Dividing by $\pi$ and multiplying by $4$:
$n+1 = 4(2k+1)$
$n+1 = 8k+4$
$n = 8k + 3$
For the smallest positive integer $n$, we substitute $k=0$, which yields $n = 3$.