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CIE 9709 2024 November Paper 32 Q3

A Level / CIE / P3

CIE 9709 2024 Nov Paper 32 Paper · Question 3

题目

Problem

The square roots of 6−8i6-8\mathrm{i} can be expressed in the Cartesian form x+iyx+\mathrm{i}y, where xx and yy are real and exact.

By first forming a quartic equation in xx or yy, find the square roots of 6−8i6-8\mathrm{i} in exact Cartesian form.

(5)
题目中文翻译

6−8i6-8\mathrm{i} 的平方根可以表示为直角坐标形式 x+iyx+\mathrm{i}y,其中 xx 和 yy 为精确的实数。

先建立一个关于 xx 或 yy 的一元四次方程,再求 6−8i6-8\mathrm{i} 的平方根,并以精确的直角坐标形式表示。

(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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先比较实部与虚部,再用乘积关系消元形成四次方程。最后配对正负号。

答题过程

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Let a square root be x+yix+y\mathrm{i}. Squaring gives

x2−y2= 6,2xy= −8.\begin{align*} x^2-y^2=&\,6,\\[4mm] 2xy=&\,-8. \end{align*}

Thus xy=−4xy=-4, so x≠0x\ne0 and y=−4/xy=-4/x. Substitution gives

x2−16x2=6.x^2-\frac{16}{x^2}=6.

The required quartic is therefore

x4−6x2−16= 0,(x2−8)(x2+2)= 0.\begin{align*} x^4-6x^2-16=&\,0,\\[4mm] (x^2-8)(x^2+2)=&\,0. \end{align*}

For real xx, x=±22x=\pm2\sqrt2. The corresponding values of yy are −2-\sqrt2 and 2\sqrt2.

Hence the square roots are

22−2 iand−22+2 i.\boxed{2\sqrt2-\sqrt2\,\mathrm{i}\quad\text{and}\quad -2\sqrt2+\sqrt2\,\mathrm{i}}.