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CIE 9709 2025 March Paper 32 Q5

A Level / CIE / P3

CIE 9709 2025 March Paper 32 Paper · Question 5

题目

Problem

The square roots of −4+65i-4+6\sqrt5\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 −4+65i-4+6\sqrt5\mathrm{i} in exact Cartesian form.

(5)
题目中文翻译

−4+65i-4+6\sqrt5\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 −4+65i-4+6\sqrt5\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,再消元形成题目要求的四次方程。乘积为正说明两个分量同号。

答题过程

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

x2−y2= −4,2xy= 65.\begin{align*} x^2-y^2=&\,-4,\\[4mm] 2xy=&\,6\sqrt5. \end{align*}

Since xy=35≠0xy=3\sqrt5\ne0, x≠0x\ne0 and y=35/xy=3\sqrt5/x. Substituting,

x2−45x2=−4.x^2-\frac{45}{x^2}=-4.

Thus the required quartic is

x4+4x2−45= 0,(x2−5)(x2+9)= 0.\begin{align*} x^4+4x^2-45=&\,0,\\[4mm] (x^2-5)(x^2+9)=&\,0. \end{align*}

Since xx is real, x=±5x=\pm\sqrt5. The corresponding values of yy are 33 and −3-3.

Therefore the two square roots are

5+3iand−5−3i.\boxed{\sqrt5+3\mathrm{i}\quad\text{and}\quad -\sqrt5-3\mathrm{i}}.