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FP2 2022 Jan Q1

A Level / Edexcel / FP2

IAL 2022 Jan Paper · Question 1

题目

Problem

1.

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(a) Express the complex number

−4−43i\begin{align*} -4 - 4\sqrt{3}\mathrm{i} \end{align*}

in the form r(cos⁡θ+isin⁡θ)r(\cos \theta + \mathrm{i} \sin \theta), where r>0r > 0 and −π<θ⩽π-\pi < \theta \leqslant \pi.

(3)

(b) Solve the equation

z3+4+43i=0\begin{align*} z^3 + 4 + 4\sqrt{3}\mathrm{i} = 0 \end{align*}

giving your answers in the form reiθr\mathrm{e}^{\mathrm{i}\theta}, where r>0r > 0 and −π<θ⩽π-\pi < \theta \leqslant \pi.

(4)
题目中文翻译

1. 本题必须展示所有解题步骤。

完全依赖计算器技术的解答不予接受。

(a) 将复数

−4−43i\begin{align*} -4-4\sqrt{3}\mathrm{i} \end{align*}

写成 r(cos⁡θ+isin⁡θ)r(\cos\theta+\mathrm{i}\sin\theta) 的形式，其中 r>0r>0 且 −π<θ⩽π-\pi<\theta\leqslant\pi。

(3)

(b) 解方程

z3+4+43i=0,\begin{align*} z^3+4+4\sqrt{3}\mathrm{i}=0, \end{align*}

将各个答案写成 reiθr\mathrm{e}^{\mathrm{i}\theta} 的形式，其中 r>0r>0 且 −π<θ⩽π-\pi<\theta\leqslant\pi。

(4)

解答

(a)

解法一

思路

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先求 modulus，再判断 argument。点 (−4,−43)(-4,-4\sqrt3) 在第三象限，所以角度应为负角

−2π3.\begin{align*} -\frac{2\pi}{3}. \end{align*}

答题过程

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Let

−4−43i=r(cos⁡θ+isin⁡θ).\begin{align*} -4-4\sqrt3\mathrm{i}=r(\cos\theta+\mathrm{i}\sin\theta). \end{align*}

The modulus is

r= (−4)2+(−43)2= 16+48= 8.\begin{align*} r =&\,\sqrt{(-4)^2+(-4\sqrt3)^2}\\[2mm] =&\,\sqrt{16+48}\\[2mm] =&\,8. \end{align*}

Also,

tan⁡θ= −43−4= 3.\begin{align*} \tan\theta =&\,\frac{-4\sqrt3}{-4}\\[2mm] =&\,\sqrt3. \end{align*}

The complex number is in the third quadrant, and the required range is −π<θ⩽π-\pi<\theta\leqslant\pi, so

θ=−2π3.\begin{align*} \theta=-\frac{2\pi}{3}. \end{align*}

Therefore

 −4−43i= 8[cos⁡(−2π3) +isin⁡(−2π3)].\begin{align*} &\,-4-4\sqrt3\mathrm{i}\\[4mm] =&\,8\bigg[\cos\Big(-\frac{2\pi}{3}\Big)\\[4mm] &\,\hspace{2pt}+\mathrm{i}\sin\Big(-\frac{2\pi}{3}\Big)\bigg]. \end{align*}

(b)

解法一

思路

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由方程得

z3=−4−43i.\begin{align*} z^3=-4-4\sqrt3\mathrm{i}. \end{align*}

利用 (a) 的模长和角度，再用 de Moivre’s theorem 取三次方根。三个根的 argument 相差 2π3\dfrac{2\pi}{3}，最后要调整到 −π<θ⩽π-\pi<\theta\leqslant\pi。

答题过程

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The equation gives

z3=−4−43i.\begin{align*} z^3=-4-4\sqrt3\mathrm{i}. \end{align*}

From part (a),

−4−43i= 8e−2π3i.\begin{align*} -4-4\sqrt3\mathrm{i} =&\,8\mathrm{e}^{-\frac{2\pi}{3}\mathrm{i}}. \end{align*}

Let

z=reiθ.\begin{align*} z=r\mathrm{e}^{\mathrm{i}\theta}. \end{align*}

Then

z3=r3e3iθ.\begin{align*} z^3=r^3\mathrm{e}^{3\mathrm{i}\theta}. \end{align*}

So

r3=8,\begin{align*} r^3=8, \end{align*}

giving

r=2.\begin{align*} r=2. \end{align*}

For the arguments,

3θ=−2π3+2kπ.\begin{align*} 3\theta=-\frac{2\pi}{3}+2k\pi. \end{align*}

Hence

θ=−2π9+2kπ3.\begin{align*} \theta=-\frac{2\pi}{9}+\frac{2k\pi}{3}. \end{align*}

The three values in the range −π<θ⩽π-\pi<\theta\leqslant\pi are

θ= −8π9,−2π9,4π9.\begin{align*} \theta =&\,-\frac{8\pi}{9},\quad -\frac{2\pi}{9},\quad \frac{4\pi}{9}. \end{align*}

Therefore

z1= 2e−8π9i,z2= 2e−2π9i,z3= 2e4π9i.\begin{align*} z_1=&\,2\mathrm{e}^{-\frac{8\pi}{9}\mathrm{i}},\\[4mm] z_2=&\,2\mathrm{e}^{-\frac{2\pi}{9}\mathrm{i}},\\[4mm] z_3=&\,2\mathrm{e}^{\frac{4\pi}{9}\mathrm{i}}. \end{align*}