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CIE 9709 2022 November Paper 33 Q5

A Level / CIE / P3

CIE 9709 2022 Nov Paper 33 Paper · Question 5

题目

Problem

(a) On a sketch of an Argand diagram, shade the region whose points represent complex numbers zz satisfying the inequalities ∣z+2∣≤2|z+2|\le2 and Im⁡z≥1\operatorname{Im}z\ge1.

(4)

(b) Find the greatest value of arg⁡z\arg z for points in the shaded region.

(2)
题目中文翻译

(a) 在阿尔冈图的草图上,用阴影表示由满足不等式 ∣z+2∣≤2|z+2|\le2 和 Im⁡z≥1\operatorname{Im}z\ge1 的复数 zz 所对应的点组成的区域。

(4)

(b) 求阴影区域内各点所对应的 arg⁡z\arg z 的最大值。

(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

第一个条件是圆心 (−2,0)(-2,0)、半径 22 的闭圆盘,第二个条件要求在水平线 y=1y=1 上方,取交集。

答题过程

展开

Write z=x+yiz=x+y\mathrm{i}. The inequalities become

(x+2)2+y2≤4,y≥1.(x+2)^2+y^2\le4,\qquad y\ge1.

Draw the circle with centre (−2,0)(-2,0) and radius 22, and the line y=1y=1. Shade the upper cap of the disc, including the circle arc and the horizontal chord.

The chord endpoints are (−2−3,1)(-2-\sqrt3,1) and (−2+3,1)(-2+\sqrt3,1).

(b)

解法一

思路

展开

区域完全在第二象限,最大辐角在最靠近负实轴的方向。沿每一水平线先取最左点,再比较高度,得到弦的左端点。

答题过程

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For 1≤y≤21\le y\le2, the leftmost point in the disc has

x=−2−4−y2.x=-2-\sqrt{4-y^2}.

At a fixed height, this point has the largest argument. Its reference-angle tangent is

y−x=y2+4−y2.\frac{y}{-x}=\frac{y}{2+\sqrt{4-y^2}}.

As yy increases from 11 to 22, the numerator increases and the positive denominator decreases, so this ratio increases. Thus the greatest argument occurs at the smallest height, y=1y=1, at (−2−3,1)(-2-\sqrt3,1).

Hence

arg⁡z= π−tan⁡−112+3= π−tan⁡−1(2−3).\begin{align*} \arg z=&\,\pi-\tan^{-1}\frac1{2+\sqrt3}\\[4mm] =&\,\pi-\tan^{-1}(2-\sqrt3). \end{align*}

Since tan⁡(π/12)=2−3\tan(\pi/12)=2-\sqrt3,

max⁡arg⁡z=11π12.\boxed{\max\arg z=\frac{11\pi}{12}}.