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CIE 9709 2023 Nov Paper 33 Q2

A Level / CIE / P3

CIE 9709 2023 Nov Paper 33 Paper · Question 2

题目

Problem

On an Argand diagram, shade the region whose points represent complex numbers zz satisfying the inequalities ∣z−1+2i∣≤∣z∣|z-1+2\mathrm{i}|\le|z| and ∣z−2∣≤1|z-2|\le1.

(5)
题目中文翻译

在 Argand 图中,用阴影标出满足不等式 ∣z−1+2i∣≤∣z∣|z-1+2\mathrm{i}|\le|z| 和 ∣z−2∣≤1|z-2|\le1 的复数 zz 所对应的点组成的区域。

(5)
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

解法一

思路

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距离不等式给出原点与 (1,−2)(1,-2) 连线的垂直平分线一侧;第二个条件是圆心 (2,0)(2,0)、半径 11 的闭圆盘。涂出两者交集。

答题过程

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Write z=x+yiz=x+y\mathrm{i}. The condition ∣z−1+2i∣≤∣z∣|z-1+2\mathrm{i}|\le|z| becomes

(x−1)2+(y+2)2≤x2+y2.(x-1)^2+(y+2)^2\le x^2+y^2.

Expanding,

−2x+4y+5≤ 0,y≤ x2−54.\begin{align*} -2x+4y+5\le&\,0,\\[4mm] y\le&\,\frac x2-\frac54. \end{align*}

Its boundary is the perpendicular bisector of the segment joining O=(0,0)O=(0,0) and P=(1,−2)P=(1,-2), whose midpoint is (1/2,−1)(1/2,-1).

The second condition is the closed disc

(x−2)2+y2≤1,(x-2)^2+y^2\le1,

with centre (2,0)(2,0) and radius 11.

Draw the circle and the line y=x/2−5/4y=x/2-5/4, mark PP, and shade the part of the disc on or below the line, including the boundaries.