Solve $z^{4} = -4 - 4\sqrt{3}\,i$, expressing your answers in the form $re^{i\theta}$, where $r > 0$ and $-\pi < \theta \le \pi$. (You do not need to list out the roots.) Hence solve $w^{4} = -1 + \sqrt{3}\,i$, expressing your answers in a similar form. Given $|p| = 2$, $\arg(p) = \dfrac{\pi}{3}$, $|q| = 7$ and $\arg(q) = -\dfrac{2\pi}{3}$, determine the modulus and argument of $\dfrac{p^{7}}{q^{3}}$. Hence express $\dfrac{p^{7}}{q^{3}}$ in the form $x + iy$, where $x, y \in \mathbb{R}$. State the smallest positive integer $n$ such that $\left(\dfrac{p^{7}}{q^{3}}\right)^{n}$ is real and negative.

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

Let $w = -4 - 4\sqrt{3}\,i$.
$|w| = \sqrt{(-4)^2 + (-4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8$
$\arg(w) = \tan^{-1}\left(\frac{-4\sqrt{3}}{-4}\right) - \pi = \frac{\pi}{3} - \pi = -\frac{2\pi}{3}$
So, $z^4 = 8e^{i\left(-\frac{2\pi}{3} + 2k\pi\right)}$ for $k \in \mathbb{Z}$.

2. Apply the fractional power to find the general expression for $z$.

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

3. Determine the values of $k$ that restrict the argument to $-\pi < \theta \le \pi$.

We require $-\pi < -\frac{\pi}{6} + \frac{k\pi}{2} \le \pi$.
For $k = -1$, $\theta = -\frac{\pi}{6} - \frac{\pi}{2} = -\frac{2\pi}{3}$
For $k = 0$, $\theta = -\frac{\pi}{6}$
For $k = 1$, $\theta = -\frac{\pi}{6} + \frac{\pi}{2} = \frac{\pi}{3}$
For $k = 2$, $\theta = -\frac{\pi}{6} + \pi = \frac{5\pi}{6}$
Hence, $k = -1, 0, 1, 2$.

1. State the roots $z$ obtained from part (a)(i) to use in this part.

From part (a)(i), we solved $z^4 = -4 - 4\sqrt{3}\,i = 8e^{-i\frac{2\pi}{3}}$.
The roots are:
$z = 8^{\frac{1}{4}} e^{i\left(\frac{-2\pi/3 + 2k\pi}{4}\right)} = 2^{\frac{3}{4}} e^{i\left(-\frac{\pi}{6} + \frac{k\pi}{2}\right)}$
for $k = -1, 0, 1, 2$.

2. Establish the algebraic relationship between the equation for $w$ and the equation for $z$.

We are asked to solve $w^4 = -1 + \sqrt{3}\,i$.
Observe how it relates to $z^4$:
$4w^4 = 4(-1 + \sqrt{3}\,i) = -4 + 4\sqrt{3}\,i$
Notice that this is the complex conjugate of $z^4$:
$4w^4 = \overline{-4 - 4\sqrt{3}\,i} = \overline{z^4}$
Taking the 4th root gives:
$w^4 = \left(\frac{\overline{z}}{\sqrt{2}}\right)^4 \implies w = \frac{1}{\sqrt{2}}\overline{z}$

3. Find the roots $w$ by applying the relationship to the roots $z$.

Applying $w = \frac{1}{\sqrt{2}}\overline{z}$ to the general root $z$:
$w = \frac{1}{\sqrt{2}} \overline{\left( 2^{\frac{3}{4}} e^{i\left(-\frac{\pi}{6} + \frac{k\pi}{2}\right)} \right)}$
$w = 2^{-\frac{1}{2}} \cdot 2^{\frac{3}{4}} e^{-i\left(-\frac{\pi}{6} + \frac{k\pi}{2}\right)} = 2^{\frac{1}{4}} e^{i\left(\frac{\pi}{6} - \frac{k\pi}{2}\right)}$
Substitute $k = 0, -1, 1, 2$ to find the four specific roots within $(-\pi, \pi]$:
For $k = 0$: $w = 2^{\frac{1}{4}} e^{i\frac{\pi}{6}}$
For $k = -1$: $w = 2^{\frac{1}{4}} e^{i\left(\frac{\pi}{6} + \frac{\pi}{2}\right)} = 2^{\frac{1}{4}} e^{i\frac{2\pi}{3}}$
For $k = 1$: $w = 2^{\frac{1}{4}} e^{i\left(\frac{\pi}{6} - \frac{\pi}{2}\right)} = 2^{\frac{1}{4}} e^{-i\frac{\pi}{3}}$
For $k = 2$: $w = 2^{\frac{1}{4}} e^{i\left(\frac{\pi}{6} - \pi\right)} = 2^{\frac{1}{4}} e^{-i\frac{5\pi}{6}}$

1. Calculate the modulus of $\dfrac{p^{7}}{q^{3}}$.

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

2. Calculate the argument of $\dfrac{p^{7}}{q^{3}}$ and find its principal value.

$\arg\left( \frac{p^{7}}{q^{3}} \right) = 7\arg(p) - 3\arg(q)$
$= 7\left(\frac{\pi}{3}\right) - 3\left(-\frac{2\pi}{3}\right)$
$= \frac{7\pi}{3} + 2\pi = \frac{13\pi}{3}$
To find the principal argument (in the range $-\pi < \theta \le \pi$), we subtract multiples of $2\pi$:
$\frac{13\pi}{3} - 4\pi = \frac{\pi}{3}$

3. Express the complex number in the form $x + iy$.

Using the modulus $r = \frac{128}{343}$ and argument $\theta = \frac{\pi}{3}$:
$\frac{p^{7}}{q^{3}} = r(\cos\theta + i\sin\theta)$
$= \frac{128}{343} \left( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right)$
$= \frac{128}{343} \left( \frac{1}{2} + i\frac{\sqrt{3}}{2} \right)$
$= \frac{64}{343} + i\frac{64\sqrt{3}}{343}$

1. Set up the condition for the expression to be real and negative using the argument from (b)(i).

From (b)(i), we have $\arg\left(\frac{p^7}{q^3}\right) = \frac{\pi}{3}$.
For a complex number to be real and negative, its principal argument must be $\pi$, or more generally, its argument must be an odd multiple of $\pi$:
$n\arg\left(\frac{p^7}{q^3}\right) = (2k + 1)\pi, \quad \text{where } k \in \mathbb{Z}$

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

Substitute the argument into the condition:
$n\left(\frac{\pi}{3}\right) = (2k + 1)\pi$
$n = 3(2k + 1)$
To find the smallest positive integer $n$, we take $k = 0$, which gives:
$n = 3(1) = 3$