复数 $z$ 满足关系式 $|z - 2\mathrm{i}| < 4$ 和 $|z + 4 - 6\mathrm{i}| \le |z - 2\mathrm{i}|$。

1. Identify the locus and boundary for the first inequality.

$|z - 2\mathrm{i}| < 4$

The boundary is $|z - 2\mathrm{i}| = 4$, which represents a circle in the Argand diagram centered at $0 + 2\mathrm{i}$ (Cartesian coordinates $(0, 2)$) with a radius of $4$.
Because the inequality is strictly less than ($<$), the region is the interior of the circle, and the boundary must be drawn as a dotted line.

2. Identify the locus and boundary for the second inequality.

$|z + 4 - 6\mathrm{i}| \le |z - 2\mathrm{i}|$
This can be rewritten as:
$|z - (-4 + 6\mathrm{i})| \le |z - 2\mathrm{i}|$

This defines the region of points closer to $-4 + 6\mathrm{i}$ (point $A(-4, 6)$) than to $2\mathrm{i}$ (point $B(0, 2)$). The boundary is the perpendicular bisector of the line segment $AB$.

Midpoint of $AB$: $\left(\frac{-4+0}{2}, \frac{6+2}{2}\right) = (-2, 4)$
Gradient of $AB$: $\frac{6-2}{-4-0} = -1$
Gradient of perpendicular bisector: $1$
Equation of boundary line: $y - 4 = 1(x - (-2)) \implies y = x + 6$

Because the inequality includes equality ($\le$), the boundary is drawn as a solid line.

3. Find the intersections of the loci to accurately determine the shaded region.

Substitute the line equation $y = x + 6$ into the Cartesian equation of the circle $x^2 + (y - 2)^2 = 16$:
$x^2 + (x + 6 - 2)^2 = 16$
$x^2 + (x + 4)^2 = 16$
$2x^2 + 8x + 16 = 16$
$2x(x + 4) = 0 \implies x = 0 \text{ or } x = -4$

When $x = 0$, $y = 6$, corresponding to the complex number $6\mathrm{i}$.
When $x = -4$, $y = 2$, corresponding to the complex number $-4 + 2\mathrm{i}$.

The line intersects the circle exactly at its top-most point $(0, 6)$ and left-most point $(-4, 2)$.

4. Synthesise the information to define the required Argand diagram sketch.

• Draw the Real ($\text{Re}$) and Imaginary ($\text{Im}$) axes.
• Draw a dotted circle centered at $(0, 2)$ passing through $(0, 6)$ and $(-4, 2)$.
• Draw a solid line passing through $(-4, 2)$ and $(0, 6)$ (with its $x$-intercept at $-6$).
• Testing the circle center $(0,2)$ in the second inequality gives $|2\mathrm{i} - (-4+6\mathrm{i})| \le 0 \implies \sqrt{32} \le 0$, which is false. Thus, shade the side of the line not containing the center.
• The correct shaded region is the minor segment of the circle bounded by the solid straight line chord between $(-4, 2)$ and $(0, 6)$ and the dotted circular arc joining them.

1. Identify the intersection points of the two boundary loci.

From part (i), the region is bounded by the circle $|z - 2\mathrm{i}| = 4$ (center $(0,2)$, radius $4$) and the perpendicular bisector $|z + 4 - 6\mathrm{i}| = |z - 2\mathrm{i}|$ of $(-4,6)$ and $(0,2)$.
The perpendicular bisector has the Cartesian equation $y - 4 = 1(x - (-2)) \implies y = x + 6$.
Substituting $y = x + 6$ into the circle's equation $x^2 + (y-2)^2 = 16$:
$x^2 + (x+4)^2 = 16 \implies 2x^2 + 8x = 0$
Solving this yields $x = 0$ and $x = -4$.
Thus, the two intersection points are $(0, 6)$ and $(-4, 2)$.

2. Determine the arguments at these intersection points.

The region lies entirely in the second quadrant. The minimum and maximum arguments occur at these two boundary points, representing the topmost and leftmost extremities of the region relative to the origin.
For the topmost point $z = 6\mathrm{i}$:
$\arg(6\mathrm{i}) = \frac{\pi}{2}$
For the leftmost point $z = -4 + 2\mathrm{i}$:
$\arg(-4 + 2\mathrm{i}) = \pi - \tan^{-1}\left(\frac{2}{4}\right) = \pi - \tan^{-1}\left(\frac{1}{2}\right)$

3. State the final range of $\arg z$, accounting for the strict inequalities.

Since the region is defined by the strict inequality $|z - 2\mathrm{i}| < 4$, the boundary arc of the circle and its intersection points are excluded from the valid region.
Thus, $\arg z$ takes all values strictly between these two bounds.
$\frac{\pi}{2} < \arg z < \pi - \tan^{-1}\left(\frac{1}{2}\right)$
(Note: $\pi - \tan^{-1}\left(\frac{1}{2}\right) \approx 2.68$ rad)

1. Deduce and describe the locus of $w$ for the sketch.

Given $\arg(w - (2 + 4\mathrm{i})) = \arg(-1 - \mathrm{i})$.
Since $-1 - \mathrm{i}$ is in the third quadrant, its argument is $-\frac{3\pi}{4}$.
The locus of $w$ is a half-line starting from $(2, 4)$ (excluding the endpoint) and extending downwards and to the left.
The gradient of this line is $\tan\left(-\frac{3\pi}{4}\right) = 1$.
The Cartesian equation of the line is $y - 4 = 1(x - 2) \Rightarrow y = x + 2$.
The locus is the portion of this line where $x < 2$.
(Students should sketch this half-line on their Argand diagram).

2. Identify the geometric relationship between the region for $z$ and the locus of $w$.

From part (i), the region for $z$ is bounded by the perpendicular bisector of $(-4, 6)$ and $(0, 2)$, which simplifies to the line $y = x + 6$. The valid region for $z$ is the segment of the circle $|z - 2\mathrm{i}| < 4$ that lies on or above this line.
Since the locus of $w$ lies on $y = x + 2$ and the boundary of $z$ lies on $y = x + 6$, the two lines have the same gradient ($m=1$) and are parallel.
The least value of $|z - w|$ corresponds to the shortest perpendicular distance between a point in the region of $z$ and a point on the half-line of $w$.

3. Calculate the exact least value of $|z - w|$.

We find the perpendicular distance between the parallel lines $y = x + 6$ and $y = x + 2$.
Take a valid point on the boundary chord of $z$, for example, its midpoint $(-2, 4)$. Note that this point is inside the circle since $|-2 + 4\mathrm{i} - 2\mathrm{i}| = |-2 + 2\mathrm{i}| = \sqrt{8} < 4$.
The line perpendicular to $y = x + 2$ passing through $(-2, 4)$ has gradient $-1$:
$y - 4 = -1(x + 2) \Rightarrow y = -x + 2$
Intersecting this with the locus of $w$ ($y = x + 2$):
$x + 2 = -x + 2 \Rightarrow 2x = 0 \Rightarrow x = 0$
When $x = 0$, $y = 2$. The point $(0, 2)$ lies on the half-line $w$ since its $x$-coordinate is less than 2.
The minimum distance is the distance between $(-2, 4)$ and $(0, 2)$:
$|z - w|_{\min} = \sqrt{(-2 - 0)^2 + (4 - 2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$