解方程 $z^4 = -4 - 4√3\mathrm{i}$，将你的答案表示为 $r\mathrm{e}^{\mathrm{i}\theta}$ 的形式，其中 $r > 0$ 且 $-\pi < \theta \le \pi$。（不需要列出所有的根。） 由此求解 $w^4 = -1 + \sqrt{3}\mathrm{i}$，将答案表示为类似的形式。 已知 $|p| = 2$，$\arg(p) = \frac{\pi}{3}$，$|q| = 7$ 且 $\arg(q) = -\frac{2\pi}{3}$，确定 $\frac{p^7}{q^3}$ 的模和辐角。由此将 $\frac{p^7}{q^3}$ 表示为 $x + \mathrm{i}y$ 的形式，其中 $x, y \in \mathbb{R}$。 确定最小的正整数 $n$，使得 $\left(\frac{p^7}{q^3}\right)^n$ 为负实数。

1. Find the modulus and argument of $-4 - 4\sqrt{3}\mathrm{i}$ to express $z^4$ in exponential form.

$|-4 - 4\sqrt{3}\mathrm{i}| = \sqrt{(-4)^2 + (-4\sqrt{3})^2} = \sqrt{16 + 48} = 8$
$\arg(-4 - 4\sqrt{3}\mathrm{i}) = -\pi + \tan^{-1}\left(\frac{4\sqrt{3}}{4}\right) = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3}$
$z^4 = 8\mathrm{e}^{\mathrm{i}\left(-\frac{2\pi}{3} + 2k\pi\right)}$

2. Solve for $z$ by taking the 4th roots, providing the general formula.

$z = \left( 8\mathrm{e}^{\mathrm{i}\left(-\frac{2\pi}{3} + 2k\pi\right)} \right)^{\frac{1}{4}}$
$z = 8^{\frac{1}{4}} \mathrm{e}^{\mathrm{i}\left(\frac{-\frac{2\pi}{3} + 2k\pi}{4}\right)}$
$z = 2^{\frac{3}{4}} \mathrm{e}^{\mathrm{i}\left(-\frac{\pi}{6} + \frac{k\pi}{2}\right)}$
To satisfy $-\pi < \theta \le \pi$, we choose $k \in \{-1, 0, 1, 2\}$.

3. Express $w^4$ in exponential form and deduce its relationship to $z^4$.

$w^4 = -1 + \sqrt{3}\mathrm{i}$
$|w^4| = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2$
$\arg(w^4) = \pi - \tan^{-1}(\sqrt{3}) = \frac{2\pi}{3}$
$w^4 = 2\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$
Observe that:
$w^4 \cdot z^4 = \left(2\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}\right)\left(8\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}\right) = 16\mathrm{e}^0 = 16$
Thus, $w^4 = \frac{16}{z^4} = \left(\frac{2}{z}\right)^4$.

4. Use the deduced relationship to find the formula for the roots of $w^4$.

The roots for $w$ are perfectly generated by $w = \frac{2}{z}$ as $z$ cycles through its 4 roots.
$w = \frac{2}{2^{\frac{3}{4}} \mathrm{e}^{\mathrm{i}\left(-\frac{\pi}{6} + \frac{k\pi}{2}\right)}}$
$w = 2^{1 - \frac{3}{4}} \mathrm{e}^{\mathrm{i}\left(0 - \left(-\frac{\pi}{6} + \frac{k\pi}{2}\right)\right)}$
$w = 2^{\frac{1}{4}} \mathrm{e}^{\mathrm{i}\left(\frac{\pi}{6} - \frac{k\pi}{2}\right)}, \quad \text{for } k \in \{-1, 0, 1, 2\}$
(Checking the arguments for $k \in \{-1, 0, 1, 2\}$ yields $\frac{2\pi}{3}, \frac{\pi}{6}, -\frac{\pi}{3}, -\frac{5\pi}{6}$, which are all within $(-\pi, \pi]$.)

1. Find the modulus of $\frac{p^7}{q^3}$.

$\left|\frac{p^7}{q^3}\right| = \frac{|p|^7}{|q|^3} = \frac{2^7}{7^3} = \frac{128}{343}$

2. Find the principal argument of $\frac{p^7}{q^3}$.

$\arg\left(\frac{p^7}{q^3}\right) = 7\arg(p) - 3\arg(q)$
$= 7\left(\frac{\pi}{3}\right) - 3\left(-\frac{2\pi}{3}\right)$
$= \frac{7\pi}{3} + 2\pi = \frac{13\pi}{3}$
Since the principal argument must be within $(-\pi, \pi]$, we subtract $4\pi$:
$\text{Principal argument} = \frac{13\pi}{3} - 4\pi = \frac{\pi}{3}$

3. Express the complex number in Cartesian form $x + \mathrm{i}y$.

$\frac{p^7}{q^3} = \frac{128}{343}\left(\cos\frac{\pi}{3} + \mathrm{i}\sin\frac{\pi}{3}\right)$
$= \frac{128}{343}\left(\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}\right)$
$= \frac{64}{343} + \mathrm{i}\frac{64\sqrt{3}}{343}$

4. Formulate the condition for the power to be real and negative.

For $\left(\frac{p^7}{q^3}\right)^n$ to be real and negative, its argument must be an odd multiple of $\pi$.
$\arg\left(\left(\frac{p^7}{q^3}\right)^n\right) = n \times \frac{\pi}{3} = (2k + 1)\pi, \quad \text{where } k \in \mathbb{Z}$

5. Solve for the smallest positive integer $n$.

To find the smallest positive integer $n$, we substitute $k = 0$:
$n \times \frac{\pi}{3} = \pi$
$n = 3$