题目

Problem

Figure 2 shows an Argand diagram for complex numbers of the form $z=x+\mathrm{i}y$ . The diagram is drawn accurately, although the scale is not shown on the axes.

Complex numbers that lie in the region $R$ , shown shaded in Figure 2, satisfy all three of the inequalities

\begin{align*} |z-15-8\mathrm{i}|\leqslant a \end{align*}

\begin{align*} 0\leqslant \arg(z+1)\leqslant b\pi \end{align*}

\begin{align*} |z+2\mathrm{i}|\geqslant |z+c\mathrm{i}| \end{align*}

where $a$ , $b$ and $c$ are real numbers.

(a) Determine the value of $a$ , the value of $b$ and the value of $c$

(3)

Given that the complex number $w$ lies in the region $R$ ,

(b) determine the exact range of possible values of $|w|$

(3)

(2)

题目中文翻译

图2是表示复数 $z=x+\mathrm{i}y$ 的阿尔冈图。图形按准确比例绘制，但坐标轴上没有标注刻度值。

阴影区域 $R$ 内的复数同时满足以下三个不等式：

\begin{align*} |z-15-8\mathrm{i}|\leqslant a,\\[3mm] 0\leqslant\arg(z+1)\leqslant b\pi,\\[3mm] |z+2\mathrm{i}|\geqslant|z+c\mathrm{i}| \end{align*}

其中 $a$ 、 $b$ 和 $c$ 均为实数。

(a) 求 $a$ 、 $b$ 和 $c$ 的值。

(3)

已知复数 $w$ 位于区域 $R$ 内。

(b) 求 $|w|$ 所有可能值的精确范围。

(3)

(2)

解答

(a)

解法一

思路

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三个不等式分别对应圆、从 $-1$ 出发的射线夹角、以及两点距离比较形成的垂直平分线。先由圆心 $(15,8)$ 和圆与实轴相切，确定半径为 $8$ ；再结合网格与圆的最左点，确定斜线角度及水平边界。

答题过程

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The inequality

\begin{align*} |z-15-8\mathrm{i}|\leqslant a \end{align*}

represents the interior and boundary of a circle with centre $(15,8)$ and radius $a$ . The circle is tangent to the real axis, so its radius equals the height of its centre:

\begin{align*} a=8 \end{align*}

The leftmost point of the circle is $(15-8,8)=(7,8)$ . The sloping boundary passes through this point and has equal horizontal and vertical steps on the grid. Its extension passes through $(-1,0)$ , as required by $\arg(z+1)$ . Its gradient is therefore

\begin{align*} \frac{8-0}{7-(-1)}=1 \end{align*}

Hence the angle is $\frac{\pi}{4}$ , giving

\begin{align*} b=\frac14 \end{align*}

The inequality

\begin{align*} |z+2\mathrm{i}|\geqslant |z+c\mathrm{i}| \end{align*}

compares distances from $(0,-2)$ and $(0,-c)$ . Its boundary is the perpendicular bisector of these two points. From the diagram this boundary is the horizontal line $y=8$ , so

\begin{align*} \frac{-2+(-c)}{2} =&\,8\\[4mm] -2-c =&\,16\\[4mm] c =&\,-18 \end{align*}

Therefore

\begin{align*} a=8,\qquad b=\frac14,\qquad c=-18 \end{align*}

(b)

解法一

思路

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$|w|$ 是点到原点的距离。最小距离在左下角点 $(7,8)$ ，最大距离在圆上沿着原点到圆心方向继续往外的点。圆心到原点距离是 $17$ ，半径是 $8$ 。

答题过程

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Every point in $R$ has $x\geqslant7$ and $y\geqslant8$ , and $(7,8)$ belongs to $R$ . Hence the minimum value of $|w|$ occurs at $(7,8)$ :

\begin{align*} |w|_{\min} =&\,\sqrt{7^2+8^2}\\[4mm] =&\,\sqrt{113} \end{align*}

The centre of the circle is $(15,8)$ , so its distance from the origin is

\begin{align*} \sqrt{15^2+8^2} =&\,17 \end{align*}

The largest possible distance from the origin to a point on this circular boundary is

\begin{align*} 17+8=25 \end{align*}

This distance is attained along the ray from the origin through the centre, at

\begin{align*} \frac{25}{17}(15,8) =&\,\Big(\frac{375}{17},\frac{200}{17}\Big) \end{align*}

This point lies on the circle, above $y=8$ and below the sloping boundary $y=x+1$ , so it belongs to $R$ . Therefore the exact range is

\begin{align*} \sqrt{113}\leqslant |w|\leqslant 25 \end{align*}

(c)

解法一

思路

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要让 $\arg w$ 最小，就是让从原点看过去的斜率 $y/x$ 尽量小。区域最低是 $y=8$ ，在这条边上越往右角越小，所以取点 $(23,8)$ 。

答题过程

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For every point in $R$ , $y\geqslant8$ and $0<x\leqslant23$ , so $y/x\geqslant8/23$ . Equality is attained at $(23,8)$ , which belongs to $R$ . Therefore

\begin{align*} \arg w_{\min} =&\,\arctan\Big(\frac{8}{23}\Big)\\[4mm] =&\,0.335\ \text{radians} \end{align*}

to 3 significant figures.

FP2 2025 June A Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程