(a) 在 Argand 图中，用阴影标出满足不等式 $|z-4-3\mathrm{i}|\le2$ 和 $\arg(z-2-\mathrm{i})\ge\frac13\pi$ 的复数 $z$ 所对应的点组成的区域。

(b) 计算此区域内各点所对应的 $\arg z$ 的最大值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

第一个条件表示圆心 $(4,3)$、半径 $2$ 的闭圆盘。第二个条件的角是从点 $(2,1)$ 出发量起的；圆盘完全位于该点的右上方，因此取斜率 $\sqrt3$ 的射线上方部分。

Write $z=x+y\mathrm{i}$. The condition $|z-4-3\mathrm{i}|\le2$ is the closed disc

Draw its boundary circle with centre $C=(4,3)$ and radius $2$.

For $\arg(z-2-\mathrm{i})\ge\pi/3$, mark $P=(2,1)$ and draw the ray from $P$ at angle $\pi/3$ to the positive real direction:

All points in the disc have $x\ge2$ and $y\ge1$. Shade the part of the disc on or above this ray, including both boundary arcs/segments.

从原点向圆作上方的切线，切点给出最大的辐角。由圆心方向角加上直角三角形中的角求解，并确认切点落在已涂区域内。

Let $O$ be the origin and $T$ the upper tangency point from $O$ to the circle. Then

Also $CT\perp OT$.

Thus

The upper tangent touches at approximately $(2.260,3.986)$. Relative to $P=(2,1)$, this point has argument approximately $1.48>\pi/3$, so it belongs to the required region.

Consequently, the greatest argument is

Thus $\boxed{\max\arg z\approx1.06\text{ radians}}$.

把从原点出发的直线写成 $y=mx$。与圆相切时，联立所得关于 $x$ 的二次方程判别式为零；取较大的斜率得到上方切线。

A line through the origin has equation $y=mx$. At an intersection with the circle,

so

For tangency, its discriminant is zero:

The larger slope gives the upper tangent. Its point of contact has

This point satisfies $y-1\ge\sqrt3(x-2)$ and hence is in the shaded region. Therefore

Hence $\boxed{\max\arg z\approx1.06\text{ radians}}$.