解方程 $i z^5 = -32$，将根写成 $re^{i\theta}$ 的形式，其中 $r>0$ 且 $-\pi<\theta\le\pi$。在阿甘特图上描绘出代表这些根的点 $P_1, P_2, P_3, P_4$ 和 $P_5$，其中 $P_1$ 代表辐角最小的根，且 $P_1 P_2 P_3 P_4 P_5$ 是按逆时针方向构成的多边形。求 $P_1 P_2 P_3 P_4 P_5$ 的面积。 在阿甘特图上清晰地描绘点 $P$ 的轨迹，其中 $P$ 代表满足 $|z-2-2i|\le 1$ 且 $\arg(z-1)=\arg(1+3i)$ 的复数 $z$。已知 $|z-2-2i|\le 1$ 且 $\arg(z-1)=\arg(1+3i)$，求 $\arg(z-3-2i)$ 的取值范围。

1. Rearrange the given equation to make $z^5$ the subject and express the right-hand side in exponential form.

$i z^5 = -32$
$z^5 = \frac{-32}{i} = -32(-i) = 32i$
To express $32i$ in exponential form $re^{i\theta}$:
The modulus is $32$, and the argument is $\frac{\pi}{2}$.
$z^5 = 32e^{i\left(\frac{\pi}{2} + 2k\pi\right)} \text{ for } k \in \mathbb{Z}$

2. Find the general expression for the roots by taking the 5th root.

$z = (32)^{1/5} e^{i\left(\frac{\frac{\pi}{2} + 2k\pi}{5}\right)}$
$z = 2 e^{i\left(\frac{\pi}{10} + \frac{2k\pi}{5}\right)}$

3. Substitute integer values for $k$ to find the 5 roots within the principal argument range $-\pi < \theta \le \pi$.

Testing consecutive integers for $k$:
For $k = 0$, $\theta = \frac{\pi}{10} \implies z = 2e^{i\frac{\pi}{10}}$
For $k = 1$, $\theta = \frac{\pi}{10} + \frac{4\pi}{10} = \frac{\pi}{2} \implies z = 2e^{i\frac{\pi}{2}}$
For $k = 2$, $\theta = \frac{\pi}{10} + \frac{8\pi}{10} = \frac{9\pi}{10} \implies z = 2e^{i\frac{9\pi}{10}}$
For $k = -1$, $\theta = \frac{\pi}{10} - \frac{4\pi}{10} = -\frac{3\pi}{10} \implies z = 2e^{-i\frac{3\pi}{10}}$
For $k = -2$, $\theta = \frac{\pi}{10} - \frac{8\pi}{10} = -\frac{7\pi}{10} \implies z = 2e^{-i\frac{7\pi}{10}}$

The 5 roots are $2e^{-i\frac{7\pi}{10}}, 2e^{-i\frac{3\pi}{10}}, 2e^{i\frac{\pi}{10}}, 2e^{i\frac{\pi}{2}}, 2e^{i\frac{9\pi}{10}}$.

4. Sketch the roots on an Argand diagram, identifying points $P_1$ to $P_5$.

The roots represent 5 equally spaced points on a circle of radius 2 centered at the origin.
Ordering the roots by smallest argument to largest:
$P_1$ represents $2e^{-i\frac{7\pi}{10}}$
$P_2$ represents $2e^{-i\frac{3\pi}{10}}$
$P_3$ represents $2e^{i\frac{\pi}{10}}$
$P_4$ represents $2e^{i\frac{\pi}{2}}$ (on the positive imaginary axis)
$P_5$ represents $2e^{i\frac{9\pi}{10}}$

(Self-drawn Argand diagram sketch showing a regular pentagon inscribed in a circle of radius 2, with vertices labelled $P_1$ to $P_5$ in an anticlockwise sequence.)

5. Calculate the area of the polygon $P_1 P_2 P_3 P_4 P_5$.

The polygon is a regular pentagon, which can be divided into 5 identical isosceles triangles extending from the origin.
For each triangle, the two equal sides have length $r = 2$, and the included angle is $\frac{2\pi}{5}$.
$\text{Area of pentagon} = 5 \times \left( \frac{1}{2} r^2 \sin\left(\frac{2\pi}{5}\right) \right)$
$\text{Area} = 5 \times \frac{1}{2} (2)^2 \sin\left(\frac{2\pi}{5}\right)$
$\text{Area} = 10\sin\left(\frac{2\pi}{5}\right) \approx 9.51 \text{ units}^2$

1. Identify the Cartesian equations for the two individual locus conditions.

Let $z = x + iy$.
The condition $|z - (2+2i)| \le 1$ represents the region inside and on the circle with centre $(2, 2)$ and radius $1$. Its Cartesian inequality is:
$(x-2)^2 + (y-2)^2 \le 1$

The condition $\arg(z-1) = \arg(1+3i)$ represents a half-line starting from $(1,0)$ with gradient $3$. Its Cartesian equation is:
$y = 3(x-1) \quad \text{for } x > 1$

2. Find the intersection of the two loci to determine the exact endpoints of the locus of $P$.

Substitute the line equation $y = 3x - 3$ into the circle's inequality:
$(x-2)^2 + (3x-3 - 2)^2 \le 1$
$(x-2)^2 + (3x-5)^2 \le 1$
$x^2 - 4x + 4 + 9x^2 - 30x + 25 \le 1$
$10x^2 - 34x + 28 \le 0$
$5x^2 - 17x + 14 \le 0$
$(5x-7)(x-2) \le 0$
This gives $1.4 \le x \le 2$.

When $x = 1.4$, $y = 3(1.4) - 3 = 1.2$.
When $x = 2$, $y = 3(2) - 3 = 3$.
The locus of $P$ is the solid line segment connecting the points $1.4 + 1.2i$ and $2 + 3i$.

3. Describe the sketch of the locus of $P$ on the Argand diagram.

On an Argand diagram, indicate the circle centre $(2,2)$ radius $1$ and the half-line from $(1,0)$ with dashed lines to show the boundaries.
The locus of $P$ is sketched by drawing a solid straight line segment from the coordinate $(1.4, 1.2)$ to the coordinate $(2, 3)$.

4. Determine the argument of $z-3-2i$ at the endpoints of the line segment locus.

Let $w = z - (3+2i)$. We need the range of $\arg(w)$ for $z$ on the line segment.
At the lower endpoint $z_1 = 1.4 + 1.2i$:
$w_1 = (1.4+1.2i) - (3+2i) = -1.6 - 0.8i$
$\arg(w_1) = -\pi + \tan^{-1}\left(\frac{0.8}{1.6}\right) = -\pi + \tan^{-1}(0.5) \approx -2.68 \text{ rad}$

At the upper endpoint $z_2 = 2 + 3i$:
$w_2 = (2+3i) - (3+2i) = -1 + i$
$\arg(w_2) = \pi - \tan^{-1}\left(\frac{1}{1}\right) = \frac{3\pi}{4} \approx 2.36 \text{ rad}$

5. Analyse the continuous variation of the argument to determine the final range.

The segment in the $w$-plane connects $w_1 = -1.6 - 0.8i$ to $w_2 = -1 + i$. Because the $y$-coordinates transition from negative ($-0.8$) to positive ($1$) whilst the $x$-coordinates remain entirely negative, the segment crosses the negative real axis (where $y = 0$).
At the point of intersection with the negative real axis, the principal argument is $\pi$.
Moving from $z_1$ to $z_2$, the argument takes all values in $(-\pi, -\pi + \tan^{-1}(0.5)]$ below the real axis, jumps to $\pi$, and takes all values in $[\frac{3\pi}{4}, \pi]$ above the real axis.

Therefore, the range of $\arg(z-3-2i)$ is:
$\left[ \frac{3\pi}{4}, \pi \right] \cup \left( -\pi, -\pi + \tan^{-1}(0.5) \right]$