复数 $z$ 的模为 2，辐角为 $\frac{\pi}{8}$。已知 $w = 1 + \mathrm{i}$。

1. Express $z^n$ in exponential form.

$z = 2\mathrm{e}^{\mathrm{i}\frac{\pi}{8}}$
$z^n = 2^n \mathrm{e}^{\mathrm{i}\frac{n\pi}{8}}$

2. Express the conjugate $w^*$ in exponential form.

$w = 1 + \mathrm{i}$
$|w| = \sqrt{1^2 + 1^2} = \sqrt{2}$
$\arg(w) = \tan^{-1}(1) = \frac{\pi}{4}$
$w = \sqrt{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$
$w^* = \sqrt{2}\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}$

3. Evaluate the quotient $\frac{z^n}{w^*}$.

$\frac{z^n}{w^*} = \frac{2^n \mathrm{e}^{\mathrm{i}\frac{n\pi}{8}}}{\sqrt{2}\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}$
$= \frac{2^n}{2^{\frac{1}{2}}} \mathrm{e}^{\mathrm{i}\left(\frac{n\pi}{8} - \left(-\frac{\pi}{4}\right)\right)}$
$= 2^{n-\frac{1}{2}} \mathrm{e}^{\mathrm{i}\left(\frac{n\pi}{8} + \frac{2\pi}{8}\right)}$
$= 2^{n-\frac{1}{2}} \mathrm{e}^{\mathrm{i}\frac{(n+2)\pi}{8}}$

1. Determine the polar form of the expression using the result from part (i).

From part (i), we establish the exponential forms:
$z = 2\mathrm{e}^{\mathrm{i}\frac{\pi}{8}} \implies z^n = 2^n \mathrm{e}^{\mathrm{i}\frac{n\pi}{8}}$
$w = 1 + \mathrm{i} = \sqrt{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{4}} \implies w^* = \sqrt{2}\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}$
$\frac{z^n}{w^*} = \frac{2^n}{\sqrt{2}} \mathrm{e}^{\mathrm{i}\left(\frac{n\pi}{8} - \left(-\frac{\pi}{4}\right)\right)} = 2^{n-\frac{1}{2}} \mathrm{e}^{\mathrm{i}\left(\frac{n\pi}{8} + \frac{\pi}{4}\right)}$

2. Set up the condition for the complex number to be real and negative.

For a complex number to be real and negative, its argument must be an odd multiple of $\pi$. Therefore, we equate the argument to $(2k + 1)\pi$ where $k \in \mathbb{Z}$:
$\frac{n\pi}{8} + \frac{\pi}{4} = (2k + 1)\pi$

3. Solve the argument equation for the smallest two positive integers $n$.

Dividing the entire equation by $\pi$ and multiplying by $8$:
$n + 2 = 8(2k + 1)$
$n = 16k + 8 - 2$
$n = 16k + 6$
Since we require $n$ to be a positive integer ($n > 0$), we take the smallest non-negative integer values for $k$:
When $k = 0$, $n = 16(0) + 6 = 6$
When $k = 1$, $n = 16(1) + 6 = 22$