在 Argand 图中，用阴影标出满足不等式 $|z-2\mathrm{i}|\le|z+2-\mathrm{i}|$ 和 $0\le\arg(z+1)\le\frac14\pi$ 的复数 $z$ 所对应的点组成的区域。

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。

模的不等式给出两点连线的垂直平分线一侧。辐角条件则从 $(-1,0)$ 出发，限定实轴方向与斜率为 $1$ 的射线之间的扇形，最后取交集。

Write $z=x+y\mathrm{i}$. Mark the points $A=(0,2)$ and $B=(-2,1)$.

The inequality $|z-2\mathrm{i}|\le|z+2-\mathrm{i}|$ means that the point is at least as close to $A$ as to $B$. Its boundary is the perpendicular bisector of $AB$:

The required side is

For $0\le\arg(z+1)\le\pi/4$, draw two rays from $(-1,0)$: the positive horizontal ray and the ray $y=x+1$ of gradient $1$.

Shade the intersection

including the boundary lines where they bound the region. The point $(-1,0)$, where the argument is undefined, is already outside the first inequality.