Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2025 June Paper 32 Q5

A Level / CIE / P3

CIE 9709 2025 June Paper 32 Paper · Question 5

题目

Problem

The square roots of −1−45i-1-4\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 −1−45i-1-4\sqrt5\mathrm{i} in exact Cartesian form.

(5)
题目中文翻译

−1−45i-1-4\sqrt5\mathrm{i} 的平方根可表示为直角坐标形式 x+iyx+\mathrm{i}y,其中 xx 和 yy 均为精确的实数。

先建立一个关于 xx 或 yy 的四次方程,再求 −1−45i-1-4\sqrt5\mathrm{i} 的平方根,答案用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

展开

按题意先建立四次方程。实虚部分别给出平方差和乘积,从乘积表示一个变量后代回,最后配对正负号。

答题过程

展开

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

x2−y2= −1,2xy= −45.\begin{align*} x^2-y^2=&\,-1,\\[4mm] 2xy=&\,-4\sqrt5. \end{align*}

Since xy=−25≠0xy=-2\sqrt5\ne0, x≠0x\ne0 and

y=−25x.y=-\frac{2\sqrt5}{x}.

Substituting into the real equation,

x2−20x2=−1.x^2-\frac{20}{x^2}=-1.

Multiplying by x2x^2 gives the required quartic:

x4+x2−20= 0,(x2−4)(x2+5)= 0.\begin{align*} x^4+x^2-20=&\,0,\\[4mm] (x^2-4)(x^2+5)=&\,0. \end{align*}

Since xx is real, x2=4x^2=4, so x=±2x=\pm2.

For x=2x=2, y=−5y=-\sqrt5; for x=−2x=-2, y=5y=\sqrt5. Therefore the square roots are

2−5 iand−2+5 i.\boxed{2-\sqrt5\,\mathrm{i}\quad\text{and}\quad -2+\sqrt5\,\mathrm{i}}.