在同一个阿根特图上，画出由 $-\frac{\pi}{4} < \arg(z + 1 + 2\text{i}) \le \frac{\pi}{4}$ 和 $|(2 + \text{i})w + 5| \le \sqrt{5}$ 所确定的轨迹。

1. Determine the centre and radius of the locus of $w$.

$|(2 + \text{i})w + 5| \le \sqrt{5}$
Factor out $(2 + \text{i})$ from the modulus:
$|2 + \text{i}| \left| w + \frac{5}{2 + \text{i}} \right| \le \sqrt{5}$
Since $|2 + \text{i}| = \sqrt{2^2 + 1^2} = \sqrt{5}$:
$\sqrt{5} \left| w + \frac{5(2 - \text{i})}{(2 + \text{i})(2 - \text{i})} \right| \le \sqrt{5}$
$\left| w + \frac{10 - 5\text{i}}{5} \right| \le 1$
$|w - (-2 + \text{i})| \le 1$
The locus of $w$ is a solid disc with centre $C(-2, 1)$ and radius $r = 1$.

2. Find the minimum argument of $w$ using the geometry of the tangent from the origin.

The minimum value of $\arg(w)$ corresponds to the upper tangent from the origin $O$ to the circle (the tangent closer to the positive imaginary axis).
The distance from the origin to the centre $C$ is $OC = \sqrt{(-2)^2 + 1^2} = \sqrt{5}$.
The argument of the centre $C$ is $\arg(C) = \pi - \tan^{-1}\left(\frac{1}{2}\right)$.
Let the angle between $OC$ and the upper tangent from $O$ be $\theta$. In the right-angled triangle formed by $O$, $C$, and the point of tangency:
$\sin \theta = \frac{r}{OC} = \frac{1}{\sqrt{5}}$
Note that $\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$ and $\tan^{-1}\left(\frac{1}{2}\right)$ describe the exact same angle, so $\theta = \tan^{-1}\left(\frac{1}{2}\right)$.
$\text{Minimum } \arg(w) = \arg(C) - \theta$
$= \pi - \tan^{-1}\left(\frac{1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right)$
$= \pi - 2 \tan^{-1}\left(\frac{1}{2}\right)$
$\approx 2.21 \text{ rad}$

1. Rewrite the inequality for $w$ to identify its locus in the Argand diagram.

The inequality $|(2 + \text{i})w + 5| \le \sqrt{5}$ can be simplified by factoring out $(2 + \text{i})$:
$|2 + \text{i}| \left| w + \frac{5}{2 + \text{i}} \right| \le \sqrt{5}$
Since $|2 + \text{i}| = \sqrt{2^2 + 1^2} = \sqrt{5}$, we have:
$\sqrt{5} \left| w + \frac{5(2 - \text{i})}{(2 + \text{i})(2 - \text{i})} \right| \le \sqrt{5}$
$\left| w + \frac{10 - 5\text{i}}{5} \right| \le 1 \implies \left| w + 2 - \text{i} \right| \le 1$
The locus of $w$ is a solid disc in the Argand diagram centered at $C(-2, 1)$ with a radius of $1$.

2. Find the Cartesian equation of the relevant boundary of the $z$-locus.

The locus of $z$ is the region between the half-lines emanating from $P(-1, -2)$ at angles $-\frac{\pi}{4}$ and $\frac{\pi}{4}$.
The upper boundary of this locus is the half-line $\arg(z + 1 + 2\text{i}) = \frac{\pi}{4}$.
This represents a ray starting from $P(-1, -2)$ with a gradient of $\tan(\frac{\pi}{4}) = 1$.
Its Cartesian equation is:
$y - (-2) = 1(x - (-1)) \implies y = x - 1 \implies x - y - 1 = 0$
where $x \ge -1$.

3. Calculate the perpendicular distance from the center of the $w$-locus to the $z$-locus boundary.

The minimum distance from $C(-2, 1)$ to the line $x - y - 1 = 0$ is the perpendicular distance:
$\text{Distance} = \frac{|(-2) - (1) - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|-4|}{\sqrt{2}} = 2\sqrt{2}$
To ensure the foot of the perpendicular lies on the valid portion of the half-line ($x \ge -1$), we find the intersection between $y = x - 1$ and the perpendicular line from $C$ ($y - 1 = -1(x + 2) \implies y = -x - 1$).
$x - 1 = -x - 1 \implies 2x = 0 \implies x = 0$
Since $x = 0 \ge -1$, the closest point on the $z$-locus to $C$ is indeed on this upper boundary at $(0, -1)$.
(Note: The distance from $C$ to the vertex $P(-1,-2)$ is $\sqrt{(-1+2)^2 + (-2-1)^2} = \sqrt{10}$, which is strictly greater than $2\sqrt{2}$.)

4. Determine the minimum value of $|z - w|$.

The value $|z - w|$ represents the distance between a point in the $z$-region and a point in the $w$-disc.
The minimum distance is the shortest distance from $C$ to the $z$-locus minus the radius of the $w$-disc:
$\text{Minimum } |z - w| = 2\sqrt{2} - 1$

1. Identify the bounding regions for $z$ and $w$ to understand the extreme directions of the vector $z-w$.

The locus of $z$ is the sector bounded by the half-lines from $(-1, -2)$ with angles $-\frac{\pi}{4}$ and $\frac{\pi}{4}$. The upper boundary is the half-line $y = x - 1$ (for $x > -1$).
The locus of $w$ is the disc $|w - (-2 + \text{i})| \le 1$, which is bounded between $x = -3$ and $x = -1$, and between $y = 0$ and $y = 2$.

2. Analyse the maximum possible argument of $z-w$.

The argument of $z-w$ is the angle the vector from $w$ to $z$ makes with the positive real axis. To maximise this angle, we must consider $z$ taking the highest possible path in its locus, which is along the upper boundary $y = x - 1$.
Since $w$ lies entirely in the region where $y \ge 0$ and $x \le -1$, for any $w$ and any $z$ on the line $y = x - 1$, the slope of the vector from $w$ to $z$ is $\frac{y_z - y_w}{x_z - x_w}$.
Since $y_z = x_z - 1$, the slope is $\frac{x_z - 1 - y_w}{x_z - x_w}$. Because $y_w - x_w > -1$ strictly for all $w$ in the disc, this slope is always strictly less than $1$.

3. Determine the limiting value to find the minimum $\theta$.

As $z$ moves towards infinity along the half-line $y = x - 1$ (i.e. $x_z \to \infty$), the slope $\frac{x_z - 1 - y_w}{x_z - x_w} \to 1$.
The vector $z-w$ becomes asymptotically parallel to the upper boundary of $z$'s locus.
Thus, $\arg(z-w)$ approaches $\frac{\pi}{4}$ from below but never reaches it.
Therefore, the strict inequality $\arg(z-w) < \frac{\pi}{4}$ holds, making the minimum value of $\theta$ exactly $\frac{\pi}{4}$.