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CIE 9709 2023 March Paper 32 Q2

A Level / CIE / P3

CIE 9709 2023 March Paper 32 Paper · Question 2

题目

Problem

(a) On an Argand diagram, shade the region whose points represent complex numbers zz satisfying the inequalities −13π≤arg⁡(z−1−2i)≤13π-\frac13\pi\le\arg(z-1-2\mathrm{i})\le\frac13\pi and Re⁡z≤3\operatorname{Re}z\le3.

(3)

(b) Calculate the least value of arg⁡z\arg z for points in the region from (a). Give your answer in radians correct to 33 decimal places.

(2)
题目中文翻译

(a) 在阿尔冈图上,用阴影表示由满足不等式 −13π≤arg⁡(z−1−2i)≤13π-\frac13\pi\le\arg(z-1-2\mathrm{i})\le\frac13\pi 和 Re⁡z≤3\operatorname{Re}z\le3 的复数 zz 所对应的点组成的区域。

(3)

(b) 求 (a) 问区域内各点所对应的 arg⁡z\arg z 的最小值。答案用弧度表示,保留 33 位小数。

(2)
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

展开

辐角是以 (1,2)(1,2) 为起点量度的,画出向右上、右下各 60∘60^\circ 的射线。与直线 x=3x=3 左侧取交集,得到三角形区域,但辐角起点本身无定义。

答题过程

展开

Write z=x+yiz=x+y\mathrm{i}. Mark P=(1,2)P=(1,2).

The argument condition describes the wedge between the rays

y−2= 3(x−1),y−2= −3(x−1),\begin{align*} y-2=&\,\sqrt3(x-1),\\[4mm] y-2=&\,-\sqrt3(x-1), \end{align*}

Both rays are restricted to x≥1x\ge1.

Draw the vertical line x=3x=3. Shade the triangular region inside the wedge and to the left of this line. Its vertices are

 (1,2), (3,2+23), (3,2−23).\begin{align*} &\,(1,2),\\[4mm] &\,(3,2+2\sqrt3),\\[4mm] &\,(3,2-2\sqrt3). \end{align*}

Include the bounding rays and the vertical segment, but exclude P=(1,2)P=(1,2) because arg⁡0\arg0 is undefined there.

(b)

解法一

思路

展开

相对于原点,最小辐角出现在三角形最右下方的顶点。可用下边界上的 y/xy/x 随 xx 递减来说明。

答题过程

展开

For each fixed xx in the region, the least argument is on the lower edge

y=2−3(x−1).y=2-\sqrt3(x-1).

Since x>0x>0, minimising the argument is equivalent to minimising y/xy/x. Along this edge,

yx=2+3x−3,\frac yx=\frac{2+\sqrt3}{x}-\sqrt3,

which decreases as xx increases. Thus the least argument is at (3,2−23)(3,2-2\sqrt3).

Therefore

min⁡arg⁡z= tan⁡−1(2−233)≈ −0.454 radians\begin{align*} \min\arg z =&\,\tan^{-1}\bigg(\frac{2-2\sqrt3}{3}\bigg)\\[4mm] \approx&\,\boxed{-0.454\text{ radians}} \end{align*}

to 3 decimal places.