解方程 $z^4 = -4 - 4\sqrt{3}\,i$，将你的答案表示为 $re^{i\theta}$ 的形式，其中 $r>0$ 且 $-\pi<\theta\le\pi$。（你不需要列出所有的根。） 由此，解方程 $w^4 = -1 + \sqrt{3}\,i$，用类似的形式表示你的答案。 已知 $|p|=2$，$\arg(p)=\frac{\pi}{3}$，$|q|=7$ 且 $\arg(q)=-\frac{2\pi}{3}$，确定 $\dfrac{p^7}{q^3}$ 的模和辐角。由此将 $\dfrac{p^7}{q^3}$ 表示为 $x+iy$ 的形式，其中 $x,y\in\mathbb{R}$。 写出使得 $\left(\dfrac{p^7}{q^3}\right)^n$ 为负实数的最小正整数 $n$。

1. Find the modulus and principal argument of the complex number $-4-4\sqrt{3}\,i$.

$|-4 - 4\sqrt{3}\,i| = \sqrt{(-4)^2 + (-4\sqrt{3})^2} = \sqrt{16 + 48} = 8$

$\arg(-4 - 4\sqrt{3}\,i) = -\left(\pi - \tan^{-1}\left(\frac{4\sqrt{3}}{4}\right)\right) = -\left(\pi - \frac{\pi}{3}\right) = -\frac{2\pi}{3}$

So, $-4 - 4\sqrt{3}\,i = 8e^{-i\frac{2\pi}{3}}$

2. Express $z^4$ using the general argument and solve for $z$.

$z^4 = 8e^{i\left(-\frac{2\pi}{3} + 2k\pi\right)}$, where $k \in \mathbb{Z}$

Taking the fourth root:
$z = 8^{\frac{1}{4}} e^{i\left(\frac{-\frac{2\pi}{3} + 2k\pi}{4}\right)}$
$z = 2^{\frac{3}{4}} e^{i\left(-\frac{\pi}{6} + \frac{k\pi}{2}\right)}$

3. Determine the specific integer values of $k$ that satisfy the principal argument range.

We require $-\pi < \theta \le \pi$:
$-\pi < -\frac{\pi}{6} + \frac{k\pi}{2} \le \pi$
$-1 < -\frac{1}{6} + \frac{k}{2} \le 1$
$-\frac{5}{6} < \frac{k}{2} \le \frac{7}{6}$
$-\frac{5}{3} < k \le \frac{7}{3}$

Since $k$ is an integer, the valid values are $k = -1, 0, 1, 2$.

1. Work out the roots of $z^4 = -4 - 4\sqrt{3}\,i$ from the earlier part to use for the deduction.

From part (a)(i), we solve $z^4 = -4 - 4\sqrt{3}\,i = 8 e^{-i\frac{2\pi}{3}}$.
Taking the 4th roots gives the modulus $|z| = 8^{1/4} = 2^{3/4}$ and arguments $\theta_k = \frac{-2\pi/3 + 2k\pi}{4}$ for $k=-1, 0, 1, 2$.
The arguments of $z$ are $-\frac{2\pi}{3}, -\frac{\pi}{6}, \frac{\pi}{3}, \frac{5\pi}{6}$.

2. Relate the new equation for $w$ to the previous equation for $z$ to apply the 'Hence' dependency.

Observe that $-1 + \sqrt{3}\,i = \frac{1}{4} (-4 + 4\sqrt{3}\,i) = \frac{1}{4} (-4 - 4\sqrt{3}\,i)^*$.
Therefore, the new equation is $w^4 = \frac{1}{4} (z^4)^* = \frac{1}{4} (z^*)^4$.
Taking the 4th root gives the algebraic relationship $w = \frac{z^*}{\sqrt{2}}$.

3. Evaluate the roots for $w$ using the relationship $w = \frac{z^*}{\sqrt{2}}$.

The modulus of $w$ is $|w| = \frac{|z^*|}{\sqrt{2}} = \frac{2^{3/4}}{2^{1/2}} = 2^{1/4}$.
The arguments of $w$ are the arguments of $z^*$, which are the negatives of the arguments of $z$:
$\arg(w) = -\arg(z) \implies \arg(w) \in \left\{ \frac{2\pi}{3}, \frac{\pi}{6}, -\frac{\pi}{3}, -\frac{5\pi}{6} \right\}$.
Hence, the roots in the required form are $w = 2^{1/4}e^{i\frac{\pi}{6}}, 2^{1/4}e^{i\frac{2\pi}{3}}, 2^{1/4}e^{-i\frac{\pi}{3}}, 2^{1/4}e^{-i\frac{5\pi}{6}}$.

1. Determine the principal argument of $\frac{p^7}{q^3}$.

$\arg\left(\frac{p^7}{q^3}\right) = 7\arg(p) - 3\arg(q)$

$= 7\left(\frac{\pi}{3}\right) - 3\left(-\frac{2\pi}{3}\right)$

$= \frac{7\pi}{3} + \frac{6\pi}{3} = \frac{13\pi}{3}$

The principal argument is $\frac{13\pi}{3} - 4\pi = \frac{\pi}{3}$.

2. Set up the condition for the power to be real and negative, and solve for the smallest positive integer $n$.

For $\left(\frac{p^7}{q^3}\right)^n$ to be real and negative, its argument must be an odd multiple of $\pi$:

$\arg\left( \left(\frac{p^7}{q^3}\right)^n \right) = n\left(\frac{\pi}{3}\right) = (2k + 1)\pi \quad \text{where } k \in \mathbb{Z}$

$n = 3(2k + 1)$

The smallest positive integer $n$ is obtained when $k=0$:

$n = 3(1) = 3$