IAL 2025 June A Q4 | 國際數學站

Figure 2

Figure 2 shows an Argand diagram for complex numbers of the form $z = x + iy$ . 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)

解答

(a)

From the diagram and the inequalities:

$|z - 15 - 8i| \le a$ is a circle with centre $15+8i$ and radius $a$ . From Figure 2, the circle is tangent to the real axis, meaning the radius is equal to the imaginary part of the centre. Therefore, $a = 8$ .
$0 \le \arg(z+1) \le b\pi$ represents the region bounded by a half-line starting at $z=-1$ . Looking at the shape of the shaded region $R$ , it lies below the line passing through $(-1,0)$ and $(7,8)$ . The slope is $\frac{8-0}{7-(-1)} = 1$ . The angle is $\arctan(1) = \frac{\pi}{4}$ . Therefore, $b = \frac{1}{4}$ .
$|z+2i| \ge |z+ci|$ represents a half-plane defined by the perpendicular bisector of the line segment between $-2i$ and $-ci$ . Since the top boundary of $R$ is a horizontal straight line passing through the centre of the circle $y=8$ , the perpendicular bisector of $-2i$ and $-ci$ must be the line $y=8$ . The distance from $y=8$ to $y=-2$ is $10$ . Thus, $-c$ must be at $y = 8 + 10 = 18$ . Therefore, $c = -18$ .

(b)

The complex number $w$ lies in $R$ . The distance from the origin $|w|$ is minimised at the intersection of the line $y=x+1$ and the horizontal line $y=8$ . This point is $(7, 8)$ .

\begin{align*} |w|_{\text{min}} = &\,\sqrt{7^2 + 8^2}\\[4mm] = &\,\sqrt{49 + 64}\\[4mm] = &\,\sqrt{113} \end{align*}

The maximum distance $|w|$ occurs at the furthest point on the circular boundary. Since $R$ is the top half of the circle $(x-15)^2 + (y-8)^2 \le 64$ , the furthest point from the origin that lies on the circle passes through the extended line from the origin to the centre $(15,8)$ . Distance to centre $= \sqrt{15^2 + 8^2} = 17$ . The furthest point on the entire circle is $17 + 8 = 25$ . This point lies in the upper half of the circle (as $y \ge 8$ ), so it is within $R$ .

\begin{align*} |w|_{\text{max}} = &\,17 + 8\\[4mm] = &\,25 \end{align*}

Thus, the exact range of possible values is:

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

(c)

The minimum value of $\arg w$ occurs at the rightmost point of the region $R$ , which is the intersection of the circle with the horizontal line $y=8$ . The circle has centre $(15, 8)$ and radius $8$ . The points on the circle where $y=8$ are $(15-8, 8)$ and $(15+8, 8)$ , which are $(7, 8)$ and $(23, 8)$ . The point that yields the minimum argument is $(23, 8)$ .

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