题目

Problem

The square roots of $3-(6\sqrt2)\mathrm{i}$ can be expressed in the Cartesian form $x+\mathrm{i}y$ , where $x$ and $y$ are real and exact.

By first forming a quartic equation, find the square roots of $3-(6\sqrt2)\mathrm{i}$ . Give your answer in exact Cartesian form.

(5)

题目中文翻译

$3-(6\sqrt2)\mathrm{i}$ 的平方根可以表示为直角坐标形式 $x+\mathrm{i}y$ ，其中 $x$ 和 $y$ 均为精确的实数。

先建立一个四次方程，再求 $3-(6\sqrt2)\mathrm{i}$ 的平方根。答案写成精确的直角坐标形式。

(5)

原卷通用作答要求

Answer all questions. Write your answers on the answer booklet. Follow the instructions on the front cover of the answer booklet. If you need additional answer paper, ask the invigilator for a continuation booklet.

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。在答题册上作答，并遵守答题册封面的指示。如需额外答题纸，请向监考人员索取续答册。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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平方后比较实部与虚部。消去一个变量，按题目要求先得到四次方程，再选取实数平方根。

答题过程

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Equating real and imaginary parts gives

\begin{align*} x^2-y^2=&\,3,\\[4mm] xy=&\,-3\sqrt2. \end{align*}

Since $x\ne0$ , substitute $y=-3\sqrt2/x$ to obtain

\begin{align*} x^2-\frac{18}{x^2}=&\,3,\\[4mm] x^4-3x^2-18=&\,0,\\[4mm] (x^2-6)(x^2+3)=&\,0. \end{align*}

Hence $x=\pm\sqrt6$ , with $y=\mp\sqrt3$ from the product condition. The two roots are

\sqrt6-\sqrt3\,\mathrm{i},\qquad -\sqrt6+\sqrt3\,\mathrm{i}.

CIE 9709 2026 June Paper 32 Q4

题目

解答

解法一

思路

答题过程