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CIE 9709 2025 November Paper 31 Q5

A Level / CIE / P3

CIE 9709 2025 Nov Paper 31 Paper · Question 5

题目

Problem

The shaded region on the Argand diagram shows points representing complex numbers zz defined by two inequalities. The shaded region is bounded by a circle and a line parallel to the imaginary axis. The boundaries of the region are included in the shaded region.

(a) Find two inequalities in terms of zz that define the shaded region.

(3)

(b) Calculate the least value of arg⁡z\arg z for points in this region.

(3)
题目中文翻译

Argand 图中的阴影区域表示由两个不等式定义的复数 zz 所对应的点。阴影区域以一个圆和一条平行于虚轴的直线为边界。阴影区域包含其边界。

(a) 求两个用 zz 表示的不等式,以定义这个阴影区域。

(3)

(b) 计算这个区域内各点所对应的 arg⁡z\arg z 的最小值。

(3)
原卷通用作答要求

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。

解答

(a)

解法一

思路

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从图上读出圆心、半径和竖直边界,并根据阴影确定不等号方向。边界包括在内,必须使用非严格不等号。

答题过程

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The circle has centre 1−2i1-2\mathrm{i} and radius 22. The shaded part lies inside the circle and to the right of Re⁡z=2\operatorname{Re}z=2.

Including the boundaries, the two inequalities are

∣z−(1−2i)∣⩽ 2,Re⁡z⩾ 2.\begin{align*} |z-(1-2\mathrm{i})|\leqslant&\,2,\\[4mm] \operatorname{Re}z\geqslant&\,2. \end{align*}

(b)

解法一

思路

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最低辐角来自竖直边界与圆的下交点。求交点后,用实部与虚部确定第四象限的负角。

答题过程

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Write z=x+yiz=x+y\mathrm{i}. On the lower circular boundary,

y=−2−4−(x−1)2,2⩽x⩽3.y=-2-\sqrt{4-(x-1)^2},\qquad 2\leqslant x\leqslant3.

As xx increases, yy increases while remaining negative. Therefore y/xy/x increases, so the least argument occurs at the lower endpoint with x=2x=2.

At this point,

(2−1)2+(y+2)2=4⟹y=−2−3.(2-1)^2+(y+2)^2=4 \quad\Longrightarrow\quad y=-2-\sqrt3.

Since the point is in the fourth quadrant,

arg⁡z=tan⁡−1(−2−32)=−1.078844….\arg z=\tan^{-1}\bigg(\frac{-2-\sqrt3}{2}\bigg) =-1.078844\ldots.

Thus the least argument is −1.08 radians\boxed{-1.08\text{ radians}}.