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CIE 9709 2022 March Paper 32 Q6

A Level / CIE / P3

CIE 9709 2022 March Paper 32 Paper · Question 6

题目

Problem

Find the complex numbers ww which satisfy the equation w2+2iw∗=1w^2+2\mathrm{i}w^*=1 and are such that Re⁡w≤0\operatorname{Re}w\leq0. Give your answers in the form x+iyx+\mathrm{i}y, where xx and yy are real.

(6)
题目中文翻译

求满足方程 w2+2iw∗=1w^2+2\mathrm{i}w^*=1 且满足 Re⁡w≤0\operatorname{Re}w\leq0 的复数 ww。答案以 x+iyx+\mathrm{i}y 的形式表示,其中 xx 和 yy 为实数。

(6)
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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设 w=x+yiw=x+y\mathrm{i},代入时共轭为 x−yix-y\mathrm{i}。比较实部、虚部后,虚部可分解成两个分支,最后用实部限制筛选。

答题过程

展开

Let w=x+yiw=x+y\mathrm{i}, with x,yx,y real. Then w∗=x−yiw^*=x-y\mathrm{i}, so

(x+yi)2+2i(x−yi)=1.(x+y\mathrm{i})^2+2\mathrm{i}(x-y\mathrm{i})=1.

Equating parts gives

x2−y2+2y= 1,2xy+2x= 2x(y+1)=0.\begin{align*} x^2-y^2+2y=&\,1,\\[4mm] 2xy+2x=&\,2x(y+1)=0. \end{align*}

If x=0x=0, the real equation gives

−y2+2y=1,(y−1)2=0,-y^2+2y=1,\qquad(y-1)^2=0,

so y=1y=1 and w=iw=\mathrm{i}.

If y=−1y=-1, the real equation gives x2−3=1x^2-3=1, so x=±2x=\pm2. The requirement Re⁡w≤0\operatorname{Re}w\le0 selects x=−2x=-2.

Therefore

w=iorw=−2−i.\boxed{w=\mathrm{i}\quad\text{or}\quad w=-2-\mathrm{i}}.