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CIE 9709 2026 June Paper 33 Q3

A Level / CIE / P3

CIE 9709 2026 June Paper 33 Paper · Question 3

题目

Problem

On an Argand diagram, shade the region which represents the complex numbers, zz, satisfying the inequalities ∣z−2+i∣≤3|z-2+\mathrm{i}|\le3 and ∣z∣≥∣z−2+2i∣|z|\ge|z-2+2\mathrm{i}|.

(5)
题目中文翻译

在 Argand 图上,用阴影标出表示满足不等式 ∣z−2+i∣≤3|z-2+\mathrm{i}|\le3 且 ∣z∣≥∣z−2+2i∣|z|\ge|z-2+2\mathrm{i}| 的复数 zz 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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第一个不等式是圆内区域。第二个比较到 00 与 2−2i2-2\mathrm{i} 的距离,分界线是两点连线的垂直平分线。

答题过程

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Write z=x+iyz=x+\mathrm{i}y. The circle is

(x−2)2+(y+1)2≤9,(x-2)^2+(y+1)^2\le9,

with centre (2,−1)(2,-1) and radius 33. The second inequality compares distances to (0,0)(0,0) and (2,−2)(2,-2). Its boundary is the perpendicular bisector through (1,−1)(1,-1), with equation y=x−2y=x-2. Squaring the distances gives

x2+y2≥ (x−2)2+(y+2)2,0≥ −4x+4y+8,y≤ x−2.\begin{align*} x^2+y^2\ge&\,(x-2)^2+(y+2)^2,\\[4mm] 0\ge&\,-4x+4y+8,\\[4mm] y\le&\,x-2. \end{align*}

Draw the circle and the line through (0,−2)(0,-2) and (2,0)(2,0) using solid boundaries. Shade the part of the closed disc on or below the line.