复数 $z$ 满足 $|z-4-3\text{i}| \le 2 \quad \text{and} \quad \pi < \arg\left[(z-4)^2\right] < 2\pi$

1. Identify the region defined by the first inequality.

The inequality $|z - 4 - 3\text{i}| \le 2$ can be rewritten as:
$|z - (4 + 3\text{i})| \le 2$
This represents the set of points inside and on the boundary of a solid circular disk with centre $C(4, 3)$ and radius $2$.

2. Determine the geometric constraint imposed by the argument inequality.

Let $\theta = \arg(z-4)$. Then $\arg\left[(z-4)^2\right] = 2\theta$ (modulo $2\pi$).
For the condition $\pi < \arg\left[(z-4)^2\right] < 2\pi$ to hold, the complex number $(z-4)^2$ must lie strictly in the lower half-plane.
This means $2\theta$ corresponds to an angle in the third or fourth quadrant, which implies $\theta$ must lie in the second or fourth quadrant:
$\theta \in \left(\frac{\pi}{2}, \pi\right) \cup \left(-\frac{\pi}{2}, 0\right)$

3. Combine both conditions to define the exact boundary of the locus.

For points in the disk $|z - (4+3\text{i})| \le 2$, the imaginary part of $z$ ranges from $3 - 2 = 1$ to $3 + 2 = 5$.
Since $\text{Im}(z) \ge 1 > 0$, the vector representing $z-4$ must point upwards, meaning its angle $\theta$ must lie in the upper half-plane, i.e., $(0, \pi)$.
Intersecting this with the quadrants found in Step 2, $\theta$ must strictly be in the second quadrant:
$\theta \in \left(\frac{\pi}{2}, \pi\right)$
This strictly requires the real part of $z-4$ to be negative:
$\text{Re}(z-4) < 0 \implies \text{Re}(z) < 4$
Thus, the required region is the strict left half of the circular disk.

1. Contextualise the locus of $z$ using the conditions from part (i).

From part (i), the condition $|z - (4+3\text{i})| \le 2$ represents a solid disk $D$ centred at $C(4,3)$ with radius 2.
For any point in $D$, $y \ge 1 > 0$, so $\arg(z-4)$ naturally lies in the interval $(0, \pi)$.
The argument condition $\pi < \arg[(z-4)^2] < 2\pi$ implies:
$\pi < 2\arg(z-4) < 2\pi \implies \frac{\pi}{2} < \arg(z-4) < \pi$
This strictly limits $z-4$ to the second quadrant, meaning $\text{Re}(z-4) < 0 \implies x < 4$.
Thus, the locus of $z$ is the left-half of $D$, excluding the vertical boundary $x=4$.

2. Calculate the maximum possible value of $|z-8|$.

$|z-8|$ represents the distance from $z$ to the point $A(8,0)$.
The distance from $A$ to the centre $C(4,3)$ is:
$|C - A| = |(4+3\text{i}) - 8| = |-4+3\text{i}| = \sqrt{(-4)^2 + 3^2} = 5$
The maximum possible distance from $A$ to the full disk $D$ is $5 + 2 = 7$.
The point achieving this maximum distance is directly opposite $A$ through $C$:
$z_{\text{max}} = A + \left(\frac{5+2}{5}\right)(C - A) = 8 + \frac{7}{5}(-4+3\text{i}) = 2.4 + 4.2\text{i}$
Since $\text{Re}(z_{\text{max}}) = 2.4 < 4$, $z_{\text{max}}$ lies well inside our restricted locus. Thus, the maximum value of $|z-8|$ is exactly $7$.

3. Determine the infimum (strict minimum limit) of $|z-8|$ constrained by the locus.

The minimum possible distance from $A$ to the full disk $D$ is $5 - 2 = 3$, which would occur at:
$z_{\text{min}} = 8 + \frac{3}{5}(-4+3\text{i}) = 5.6 + 1.8\text{i}$
Since $\text{Re}(z_{\text{min}}) = 5.6 \ge 4$, this unconstrained minimum is outside the locus.
For points restricted to the locus, the square of the distance to $A(8,0)$ is $(x-8)^2 + y^2$.
Because $x < 4$, we have $x-8 < -4 \implies (x-8)^2 > 16$.
Additionally, for all points inside the disk $D$, the minimum possible value for $y$ is $1$.
As $z$ gets arbitrarily close to $4+\text{i}$ (the bottom-right limiting boundary of the locus), the distance approaches:
$\sqrt{(4-8)^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17}$
Because the boundary $x=4$ is strictly excluded, $|z-8|$ must be strictly greater than $\sqrt{17}$.

4. State the final range of values.

Combining the strict lower bound and the achieved maximum distance, the range of values is:
$\sqrt{17} < |z-8| \le 7$

1. Identify that the maximum argument corresponds to the tangent from $z=8$ to the circular locus.

The locus of $z$ is a region within the circle $|z-(4+3\text{i})| \le 2$. The expression $\arg(z-8)$ represents the angle made by the vector from $P(8,0)$ to a point $z$ in the locus with the positive real axis.
To maximise this angle, we must consider the tangent line from $P(8,0)$ to the circle that gives the largest angle, touching the circle at a point $T$.

2. Calculate the relevant angles in the right-angled triangle formed by the centre of the circle, the point $z=8$, and the point of tangency.

Let $C(4,3)$ be the centre of the circle. In the right-angled $\triangle PTC$, the distance from $P(8,0)$ to $C(4,3)$ is:
$PC = |4+3\text{i} - 8| = |-4+3\text{i}| = \sqrt{(-4)^2 + 3^2} = 5$
The radius of the circle is $CT = 2$.
The angle $\angle CPT$ between the line $PC$ and the tangent line $PT$ is given by:
$\sin(\angle CPT) = \frac{CT}{PC} = \frac{2}{5} \implies \angle CPT = \arcsin\left(\frac{2}{5}\right) \approx 0.4115 \text{ rad}$

3. Find the angle of the line $PC$ and combine it with $\angle CPT$ to find the maximum argument.

The angle of the vector $\vec{PC} = -4+3\text{i}$ from the positive real axis is:
$\pi - \arctan\left(\frac{3}{4}\right)$
To maximise the argument, the tangent $PT$ must be rotated further counter-clockwise from $PC$ relative to $P$. Thus, we add $\angle CPT$ to the angle of $PC$:
$\max \arg(z-8) = \pi - \arctan\left(\frac{3}{4}\right) + \angle CPT$
$\max \arg(z-8) = \pi - \arctan\left(\frac{3}{4}\right) + \arcsin\left(\frac{2}{5}\right)$
$\max \arg(z-8) \approx \pi - 0.6435 + 0.4115 \approx 2.91 \text{ rad}$
(Note: The point of tangency $T$ has $x \approx 3.54 < 4$, which strictly satisfies the locus condition $\pi < \arg\left[(z-4)^2\right] < 2\pi$.)