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IAL 2025 June Q1

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 1

题目

Problem

(a) Express −22−26 i-2\sqrt2-2\sqrt6\,\mathrm{i} in the form reiθr\mathrm{e}^{\mathrm{i}\theta} where −π<θ⩽π-\pi<\theta\leqslant\pi

(3)

(b) Hence solve

z5=−22−26 i\begin{align*} z^5=-2\sqrt2-2\sqrt6\,\mathrm{i} \end{align*}

Give your answers in the form p eiθ\sqrt{p}\,\mathrm{e}^{\mathrm{i}\theta} where p∈Z+p\in\mathbb{Z}^+ and −π<θ⩽π-\pi<\theta\leqslant\pi

(3)

解答

(a)

解法一

思路

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先求模长，再判断象限。实部和虚部都为负，所以点在第三象限；但题目要求 −π<θ⩽π-\pi<\theta\leqslant\pi，因此用负角表示更自然。

答题过程

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Let

w=−22−26 i\begin{align*} w=-2\sqrt2-2\sqrt6\,\mathrm{i} \end{align*}

The modulus is

r=(−22)2+(−26)2=8+24=42\begin{align*} r ={}& \sqrt{(-2\sqrt2)^2+(-2\sqrt6)^2}\\[3mm] ={}& \sqrt{8+24}\\[3mm] ={}& 4\sqrt2 \end{align*}

For the argument,

tan⁡α=2622=3\begin{align*} \tan\alpha ={}& \frac{2\sqrt6}{2\sqrt2}\\[2mm] ={}& \sqrt3 \end{align*}

so

α=π3\begin{align*} \alpha=\frac{\pi}{3} \end{align*}

Since ww is in the third quadrant and −π<θ⩽π-\pi<\theta\leqslant\pi,

θ=−π+π3=−2π3\begin{align*} \theta ={}& -\pi+\frac{\pi}{3}\\[2mm] ={}& -\frac{2\pi}{3} \end{align*}

Therefore

−22−26 i=42 e−2π3i\begin{align*} \boxed{ -2\sqrt2-2\sqrt6\,\mathrm{i} =4\sqrt2\,\mathrm{e}^{-\frac{2\pi}{3}\mathrm{i}} } \end{align*}

(b)

解法一

思路

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由 (a)，右边的模长是 42=25/24\sqrt2=2^{5/2}。开五次方后模长是 2\sqrt2。辐角要加上 2kπ2k\pi 后再除以 55，最后选出满足 −π<θ⩽π-\pi<\theta\leqslant\pi 的五个角。

答题过程

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From part (a),

z5=42 e−2π3i\begin{align*} z^5 ={}& 4\sqrt2\,\mathrm{e}^{-\frac{2\pi}{3}\mathrm{i}} \end{align*}

Thus

z=(42)1/5ei(−2π3+2kπ5)=2 ei(−2π15+2kπ5)\begin{align*} z ={}& (4\sqrt2)^{1/5} \mathrm{e}^{\mathrm{i}\left( \frac{-\frac{2\pi}{3}+2k\pi}{5} \right)}\\[4mm] ={}& \sqrt2\, \mathrm{e}^{\mathrm{i}\left( -\frac{2\pi}{15}+\frac{2k\pi}{5} \right)} \end{align*}

For k=0,1,2,3,4k=0,1,2,3,4, the arguments are

−2π15,4π15,10π15,16π15,22π15\begin{align*} -\frac{2\pi}{15},\quad \frac{4\pi}{15},\quad \frac{10\pi}{15},\quad \frac{16\pi}{15},\quad \frac{22\pi}{15} \end{align*}

Convert those outside −π<θ⩽π-\pi<\theta\leqslant\pi:

16π15−2π=−14π15,22π15−2π=−8π15\begin{align*} \frac{16\pi}{15}-2\pi ={}& -\frac{14\pi}{15},\\ \frac{22\pi}{15}-2\pi ={}& -\frac{8\pi}{15} \end{align*}

Therefore the five roots are

2 e−14π15i,2 e−8π15i,2 e−2π15i,2 e4π15i,2 e2π3i\begin{align*} \boxed{ \sqrt2\,\mathrm{e}^{-\frac{14\pi}{15}\mathrm{i}}, \quad \sqrt2\,\mathrm{e}^{-\frac{8\pi}{15}\mathrm{i}}, \quad \sqrt2\,\mathrm{e}^{-\frac{2\pi}{15}\mathrm{i}}, \quad \sqrt2\,\mathrm{e}^{\frac{4\pi}{15}\mathrm{i}}, \quad \sqrt2\,\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}} } \end{align*}