解方程 $z^5 + 16 = 16\sqrt{3}\, i$，将根表示为 $r e^{i\theta}$ 的形式，其中 $r>0$ 且 $-\pi < \theta \le \pi$。在阿根特图上表示这些根。 点 $A$ 和 $B$ 表示具有最小的两个正辐角的根。点 $P$ 是 $AB$ 的中点，表示复数 $w$。

1. State the two roots $z_A$ and $z_B$ with the smallest positive arguments from the earlier unlabelled part.

From the equation $z^5 = -16 + 16\sqrt{3}i = 32e^{i\frac{2\pi}{3}}$, the roots are:
$z = 2e^{i\left(\frac{2\pi}{15} + \frac{2k\pi}{5}\right)}, \quad k \in \mathbb{Z}$
The two roots with the smallest positive arguments correspond to $k=0$ and $k=1$:
$z_A = 2e^{i\frac{2\pi}{15}}, \quad z_B = 2e^{i\frac{8\pi}{15}}$

2. Set up the expression for the midpoint $w$.

The midpoint $w$ of $A$ and $B$ is given by $w = \frac{z_A + z_B}{2}$:
$w = \frac{2e^{i\frac{2\pi}{15}} + 2e^{i\frac{8\pi}{15}}}{2}$
$w = e^{i\frac{2\pi}{15}} + e^{i\frac{8\pi}{15}}$

3. Factor out the exponential of the mean argument to simplify the expression for $w$.

Factor out $e^{i\frac{5\pi}{15}}$ (where $\frac{5\pi}{15} = \frac{\pi}{3}$ is the average of the two arguments $\frac{2\pi}{15}$ and $\frac{8\pi}{15}$):
$w = e^{i\frac{5\pi}{15}} \left( e^{i\left(\frac{2\pi}{15} - \frac{5\pi}{15}\right)} + e^{i\left(\frac{8\pi}{15} - \frac{5\pi}{15}\right)} \right)$
$w = e^{i\frac{\pi}{3}} \left( e^{-i\frac{3\pi}{15}} + e^{i\frac{3\pi}{15}} \right)$
$w = e^{i\frac{\pi}{3}} \left( e^{-i\frac{\pi}{5}} + e^{i\frac{\pi}{5}} \right)$
Using Euler's formula $e^{i\theta} + e^{-i\theta} = 2\cos\theta$:
$w = 2\cos\left(\frac{\pi}{5}\right)e^{i\frac{\pi}{3}}$

4. Identify the exact modulus and argument of $w$.

Since the angle $\frac{\pi}{5}$ is in the first quadrant, $2\cos\left(\frac{\pi}{5}\right) > 0$. Thus, $w$ is perfectly expressed in the valid polar form $re^{i\theta}$ where $r > 0$.
Therefore, the exact modulus and argument are:
$|w| = 2\cos\left(\frac{\pi}{5}\right)$
$\arg(w) = \frac{\pi}{3}$

1. Recall or re-derive the exponential form of $w$ from part (i).

From part (i), the two roots of $z^5 = 32e^{i\frac{2\pi}{3}}$ with the smallest positive arguments are $A = 2 e^{i\frac{2\pi}{15}}$ and $B = 2 e^{i\frac{8\pi}{15}}$.
The midpoint $P$ represents $w$:
$w = \frac{A+B}{2} = e^{i\frac{2\pi}{15}} + e^{i\frac{8\pi}{15}}$
Factoring out $e^{i\frac{5\pi}{15}}$ (which simplifies to $e^{i\frac{\pi}{3}}$):
$w = e^{i\frac{\pi}{3}}\left(e^{-i\frac{3\pi}{15}} + e^{i\frac{3\pi}{15}}\right) = e^{i\frac{\pi}{3}}\left(e^{-i\frac{\pi}{5}} + e^{i\frac{\pi}{5}}\right) = 2\cos\left(\frac{\pi}{5}\right)e^{i\frac{\pi}{3}}$
Thus, $|w| = 2\cos\left(\frac{\pi}{5}\right)$ and $\arg(w) = \frac{\pi}{3}$.

2. Set up the expression for $w^n = k$.

Since $w$ is an $n^{\text{th}}$ root of $k$, we have $w^n = k$.
Substitute the exponential form of $w$:
$k = \left( 2\cos\left(\frac{\pi}{5}\right) e^{i\frac{\pi}{3}} \right)^n = 2^n \cos^n\left(\frac{\pi}{5}\right) e^{i\frac{n\pi}{3}}$
Expressed in terms of real and imaginary parts:
$k = 2^n \cos^n\left(\frac{\pi}{5}\right) \left[ \cos\left(\frac{n\pi}{3}\right) + i\sin\left(\frac{n\pi}{3}\right) \right]$

3. Apply the condition that $k$ is a real number to find the least possible positive integer $n$.

For $k$ to be a purely real number, its imaginary part must be zero:
$\sin\left(\frac{n\pi}{3}\right) = 0$
This means the angle $\frac{n\pi}{3}$ must be a multiple of $\pi$, giving:
$\frac{n\pi}{3} = m\pi \implies n = 3m$ for some integer $m$.
Since $n$ is required to be a positive integer, the least possible value is obtained when $m = 1$:
$n = 3$

4. Calculate the corresponding real value of $k$ in trigonometric form.

Substitute $n = 3$ back into the equation for $k$:
$k = 2^3 \cos^3\left(\frac{\pi}{5}\right) \left[ \cos(\pi) + i\sin(\pi) \right]$
$k = 8 \cos^3\left(\frac{\pi}{5}\right) (-1 + 0) = -8 \cos^3\left(\frac{\pi}{5}\right)$
(The expression is left in terms of the trigonometric function $\cos\left(\frac{\pi}{5}\right)$ to satisfy the requirement to leave the answer in trigonometric form.)

1. Identify the roots to be plotted

From solving $z^5 = -16 + 16\sqrt{3}i = 32e^{i\frac{2\pi}{3}}$, the five roots are:
$z_1 = 2e^{i\frac{2\pi}{15}}$
$z_2 = 2e^{i\frac{8\pi}{15}}$
$z_3 = 2e^{i\frac{14\pi}{15}}$
$z_4 = 2e^{-i\frac{4\pi}{15}}$
$z_5 = 2e^{-i\frac{2\pi}{3}}$

2. Describe the points on the Argand diagram

Draw the real (Re) and imaginary (Im) axes.
Plot five points forming a regular pentagon on a circle of radius $2$ centered at the origin.
Mark the points at angles (measured counter-clockwise from the positive Re axis): $\frac{2\pi}{15}$, $\frac{8\pi}{15}$, $\frac{14\pi}{15}$, $-\frac{4\pi}{15}$, and $-\frac{2\pi}{3}$.