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CIE 9231 2025 Nov Paper 22 Q2

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 22 Paper · Question 2

题目

Problem

Find the roots of the equation z4=8−8i3z^4 = 8 - 8\mathrm{i}\sqrt{3}, giving your answers in the form reiθre^{\mathrm{i}\theta}, where r>0r > 0 and −π≤θ<π-\pi \leq \theta < \pi.

[5]
题目中文翻译

求方程 z4=8−8i3z^4 = 8 - 8\mathrm{i}\sqrt{3} 的根。答案写成 reiθre^{\mathrm{i}\theta} 的形式,其中 r>0r > 0 且 −π≤θ<π-\pi \leq \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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先把右边写成指数形式。四次根的辐角相差 π2\frac{\pi}{2},最后把四个辐角都移到指定区间。

答题过程

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The number 8−8i38-8\mathrm{i}\sqrt3 has modulus 1616 and argument −π/3-\pi/3. Its fourth roots have modulus 22 and arguments −π/12+kπ/2-\pi/12+k\pi/2 for k∈Zk\in\mathbb Z. Choosing four distinct arguments in [−π,π)[-\pi,\pi) gives

z1= 2e−iπ/12,z2=2e5iπ/12,z3= 2e11iπ/12,z4=2e−7iπ/12.\begin{align*} z_1=&\,2e^{-\mathrm{i}\pi/12},\qquad z_2=2e^{5\mathrm{i}\pi/12},\\ z_3=&\,2e^{11\mathrm{i}\pi/12},\qquad z_4=2e^{-7\mathrm{i}\pi/12}. \end{align*}