Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9231 2024 June Paper 22 Q1

A Level / CIE / FP2

CIE 9231 2024 June Paper 22 Paper · Question 1

题目

Problem

Find the roots of the equation

z3=−1083+108iz^3=-108\sqrt{3}+108\mathrm{i}

giving your answers in the form r(cos⁡θ+isin⁡θ)r(\cos\theta+\mathrm{i}\sin\theta), where r>0r>0 and 0<θ<2π0<\theta<2\pi.

[5]
题目中文翻译

求方程

z3=−1083+108iz^3=-108\sqrt{3}+108\mathrm{i}

的根,并将答案写成 r(cos⁡θ+isin⁡θ)r(\cos\theta+\mathrm{i}\sin\theta) 的形式,其中 r>0r>0 且 0<θ<2π0<\theta<2\pi。

(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

先求右边的模和辐角,再用 de Moivre 定理求三次根。辐角须落在 0<θ<2π0<\theta<2\pi,相邻根相差 2π3\frac{2\pi}{3}。

答题过程

展开

The number −1083+108i-108\sqrt3+108\mathrm{i} has modulus 1082(3+1)=216\sqrt{108^2(3+1)}=216 and argument 5π/65\pi/6. Its cube roots have modulus 2161/3=6216^{1/3}=6 and arguments 5π18+2kπ3\frac{5\pi}{18}+\frac{2k\pi}{3} for integer kk. In the required range the roots are

z1= 6(cos⁡5π18+isin⁡5π18),z2= 6(cos⁡17π18+isin⁡17π18),z3= 6(cos⁡29π18+isin⁡29π18).\begin{align*} z_1=&\,6\bigg(\cos\frac{5\pi}{18}+\mathrm{i}\sin\frac{5\pi}{18}\bigg),\\ z_2=&\,6\bigg(\cos\frac{17\pi}{18}+\mathrm{i}\sin\frac{17\pi}{18}\bigg),\\ z_3=&\,6\bigg(\cos\frac{29\pi}{18}+\mathrm{i}\sin\frac{29\pi}{18}\bigg). \end{align*}