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CIE 9709 2024 June Paper 33 Q3

A Level / CIE / P3

CIE 9709 2024 June Paper 33 Paper · Question 3

题目

Problem

The square roots of 24−7i24-7\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 24−7i24-7\mathrm{i} in exact Cartesian form.

(5)
题目中文翻译

24−7i24-7\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 为实数,且取精确值。

先建立一个关于 xx 或 yy 的四次方程,再求 24−7i24-7\mathrm{i} 的平方根,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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设平方根为 u+viu+v\mathrm{i},平方后比较实部、虚部。消元得到关于 u2u^2 的二次方程,再利用 uv<0uv<0 配对两个符号。

答题过程

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Let a square root be u+viu+v\mathrm{i}, where u,vu,v are real. Then

(u+vi)2=u2−v2+2uvi=24−7i.(u+v\mathrm{i})^2=u^2-v^2+2uv\mathrm{i}=24-7\mathrm{i}.

Equating real and imaginary parts,

u2−v2= 24,2uv= −7.\begin{align*} u^2-v^2=&\,24,\\[4mm] 2uv=&\,-7. \end{align*}

Since u≠0u\ne0, v=−7/(2u)v=-7/(2u). Substitution gives

u2−494u2= 24,4u4−96u2−49= 0.\begin{align*} u^2-\frac{49}{4u^2}=&\,24,\\[4mm] 4u^4-96u^2-49=&\,0. \end{align*}

Solving the quadratic in u2u^2,

u2=96±962+16(49)8=96±1008.u^2=\frac{96\pm\sqrt{96^2+16(49)}}8 =\frac{96\pm100}{8}.

The value u2=−1/2u^2=-1/2 is impossible for real uu. Thus u2=49/2u^2=49/2, so u=±72/2u=\pm7\sqrt2/2.

Using 2uv=−72uv=-7 to pair the signs, the two square roots are

722−22iand−722+22i.\boxed{\frac{7\sqrt2}{2}-\frac{\sqrt2}{2}\mathrm{i} \quad\text{and}\quad -\frac{7\sqrt2}{2}+\frac{\sqrt2}{2}\mathrm{i}}.