一个复数 $z$ 由 $z = r \mathrm{e}^{\mathrm{i}\frac{\pi}{n}}$ 给出，其中 $0 < r < 1$ 且 $n$ 为满足 $n \ge 3$ 的正整数。数 $1$，$z$，$z^2$，...，$z^n$ 可分别由阿尔冈图中的点 $P_0$，$P_1$，$P_2$，...，$P_n$ 表示。由点 $P_0$，$P_1$，$P_2$，...，$P_n$ 构成的 $(n+1)$ 边形称为由 $z$ 生成的 $(n+1)$ 边形。下方的阿尔冈图（未按比例绘制）示出了一个由 $z$ 生成的 $(3+1)$ 边形示例。 对于部分 (a) 至 (c)，设 $z = \frac{1}{4}(1 + \sqrt{3}\mathrm{i})$。

1. Calculate the modulus $r$ and the argument $\theta$ of $z$.

$r = |z| = \sqrt{\left(\frac{1}{4}\right)^2 + \left(\frac{\sqrt{3}}{4}\right)^2} = \sqrt{\frac{1}{16} + \frac{3}{16}} = \frac{1}{2}$

Since $z$ is in the first quadrant, its argument is:
$\theta = \arg(z) = \tan^{-1}\left(\frac{\sqrt{3}/4}{1/4}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$

2. Express $z$ in exponential form.

$z = \frac{1}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$

1. State the exponential form of $z$ from part (a) and compute $z^2$.

From part (a), $z = \frac{1}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$.

$z^2 = \left(\frac{1}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}\right)^2$
$z^2 = \left(\frac{1}{2}\right)^2 \mathrm{e}^{\mathrm{i}\left(\frac{\pi}{3} \times 2\right)}$
$z^2 = \frac{1}{4}\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$

2. Compute $z^3$ using the same index laws.

$z^3 = \left(\frac{1}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}\right)^3$
$z^3 = \left(\frac{1}{2}\right)^3 \mathrm{e}^{\mathrm{i}\left(\frac{\pi}{3} \times 3\right)}$
$z^3 = \frac{1}{8}\mathrm{e}^{\mathrm{i}\pi}$

1. Find the area of triangle $OP_0P_1$ using the modulus and argument of $z$.

From part (a), $z = \frac{1}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$.
In the Argand diagram, $P_0$ corresponds to $z^0 = 1$ and $P_1$ corresponds to $z$.
Thus, the lengths $OP_0 = 1$ and $OP_1 = |z| = \frac{1}{2}$.
The angle between $OP_0$ and $OP_1$ is $\arg(z) = \frac{\pi}{3}$.
$\text{Area of } \triangle OP_0P_1 = \frac{1}{2}(OP_0)(OP_1)\sin\left(\frac{\pi}{3}\right)$
$= \frac{1}{2}(1)\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{8}$

2. Deduce that the area of the polygon is the sum of the areas of three triangles meeting at the origin.

The $(3+1)$-polygon is a quadrilateral with vertices $P_0, P_1, P_2, P_3$.
From part (b), $P_3$ corresponds to $z^3 = \frac{1}{8}\mathrm{e}^{\mathrm{i}\pi}$, which lies on the negative real axis.
Since $P_0$ lies on the positive real axis, the origin $O$ lies exactly on the line segment connecting $P_3$ and $P_0$.
Therefore, the total area of the polygon is simply the sum of the areas of the constituent triangles: $\triangle OP_0P_1$, $\triangle OP_1P_2$, and $\triangle OP_2P_3$.

3. Calculate the areas of the remaining triangles and sum them up to find the exact area of the polygon.

For $\triangle OP_1P_2$:
$OP_1 = \frac{1}{2}$, $OP_2 = |z^2| = \frac{1}{4}$, and $\angle P_1OP_2 = \arg(z^2) - \arg(z) = \frac{\pi}{3}$.
$\text{Area of } \triangle OP_1P_2 = \frac{1}{2}\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{32}$

For $\triangle OP_2P_3$:
$OP_2 = \frac{1}{4}$, $OP_3 = |z^3| = \frac{1}{8}$, and $\angle P_2OP_3 = \arg(z^3) - \arg(z^2) = \frac{\pi}{3}$.
$\text{Area of } \triangle OP_2P_3 = \frac{1}{2}\left(\frac{1}{4}\right)\left(\frac{1}{8}\right)\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{128}$

$\text{Exact Area} = \frac{\sqrt{3}}{8} + \frac{\sqrt{3}}{32} + \frac{\sqrt{3}}{128}$
$= \frac{16\sqrt{3} + 4\sqrt{3} + \sqrt{3}}{128} = \frac{21\sqrt{3}}{128}$

1. Determine the area of the $k$-th triangle $OP_{k-1}P_k$.

Let the vertices of the $(n+1)$-polygon be $P_0, P_1, \dots, P_n$. The point $P_k$ represents the complex number:
$z^k = \left(\frac{1}{2} \mathrm{e}^{\mathrm{i}\frac{\pi}{n}}\right)^k = \left(\frac{1}{2}\right)^k \mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}$
The distance from the origin $O$ to $P_k$ is $OP_k = |z^k| = \left(\frac{1}{2}\right)^k$.
The angle between the segments $OP_{k-1}$ and $OP_k$ is the difference in their arguments:
$\arg(z^k) - \arg(z^{k-1}) = \frac{k\pi}{n} - \frac{(k-1)\pi}{n} = \frac{\pi}{n}$
The area of triangle $OP_{k-1}P_k$ is:
$\text{Area} = \frac{1}{2} (OP_{k-1})(OP_k) \sin\left(\frac{\pi}{n}\right) = \frac{1}{2} \left(\frac{1}{2}\right)^{k-1} \left(\frac{1}{2}\right)^k \sin\left(\frac{\pi}{n}\right) = \left(\frac{1}{4}\right)^k \sin\left(\frac{\pi}{n}\right)$

2. Express the total area of the polygon as a sum of the triangular areas.

The point $P_0 = 1$ lies on the positive real axis, and $P_n = z^n = \left(\frac{1}{2}\right)^n \mathrm{e}^{\mathrm{i}\pi} = -\left(\frac{1}{2}\right)^n$ lies on the negative real axis.
Since the line segment connecting $P_n$ and $P_0$ passes along the real axis directly through the origin $O$, the $(n+1)$-polygon is exactly partitioned into the $n$ triangles $OP_0P_1, OP_1P_2, \dots, OP_{n-1}P_n$.
The total area of the polygon is the sum of the areas of these $n$ triangles:
$\text{Total Area} = \sum_{k=1}^n \left(\frac{1}{4}\right)^k \sin\left(\frac{\pi}{n}\right) = \sin\left(\frac{\pi}{n}\right) \left[ \frac{1}{4} + \left(\frac{1}{4}\right)^2 + \dots + \left(\frac{1}{4}\right)^n \right]$

3. Evaluate the geometric series to obtain the final expression and determine the constants $a$ and $b$.

The terms inside the square bracket form a finite geometric series with $n$ terms, a first term of $u_1 = \frac{1}{4}$, and a common ratio of $r = \frac{1}{4}$.
Using the formula for the sum of a finite geometric series:
$\sum_{k=1}^n \left(\frac{1}{4}\right)^k = \frac{\frac{1}{4}\left(1 - \left(\frac{1}{4}\right)^n\right)}{1 - \frac{1}{4}} = \frac{\frac{1}{4}\left(1 - \left(\frac{1}{4}\right)^n\right)}{\frac{3}{4}} = \frac{1}{3}\left(1 - \left(\frac{1}{4}\right)^n\right)$
Substituting this back, the total area of the polygon is:
$\frac{1}{3}\left(1 - \left(\frac{1}{4}\right)^n\right) \sin\left(\frac{\pi}{n}\right)$
Comparing this to the required form $a(1 - b^n)\sin\frac{\pi}{n}$, we determine that $a = \frac{1}{3}$ and $b = \frac{1}{4}$.