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CIE 9231 2025 June Paper 21 Q1

A Level / CIE / FP2

CIE 9231 2025 June Paper 21 Paper · Question 1

题目

Problem

Find the roots of the equation z3=27−27iz^3 = 27 - 27\mathrm{i}, giving your answers in the form reiθre^{\mathrm{i}\theta}, where r>0r > 0 and −π≤θ<π-\pi \le \theta < \pi.

[5]
题目中文翻译

求方程 z3=27−27iz^3 = 27 - 27\mathrm{i} 的根,答案写成 reiθre^{\mathrm{i}\theta} 的形式,其中 r>0r > 0 且 −π≤θ<π-\pi \le \theta < \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

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先把右边化为指数形式。取三次根时,模长开立方,辐角加上 2kπ2k\pi 再除以 33。

答题过程

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We have 27−27i=272 e−iπ/427-27\mathrm{i}=27\sqrt2\,e^{-\mathrm{i}\pi/4}. Its cube roots have modulus 3 21/63\,2^{1/6} and arguments −π12+2kπ3-\frac{\pi}{12}+\frac{2k\pi}{3}, where kk is an integer. The three distinct roots with arguments in [−π,π)[-\pi,\pi) are

z1= 3 21/6e−iπ/12,z2= 3 21/6e7iπ/12,z3= 3 21/6e−3iπ/4.\begin{align*} z_1=&\,3\,2^{1/6}e^{-\mathrm{i}\pi/12},\\ z_2=&\,3\,2^{1/6}e^{7\mathrm{i}\pi/12},\\ z_3=&\,3\,2^{1/6}e^{-3\mathrm{i}\pi/4}. \end{align*}