题目

Problem

1. Solve the equation

\begin{align*} z^5-32\mathrm{i}=&\,0. \end{align*}

giving each answer in the form $r\mathrm{e}^{\mathrm{i}\theta}$ where $0 < \theta < 2\pi$

(4)

题目中文翻译

1. 解方程

\begin{align*} z^5-32\mathrm{i}=&\,0. \end{align*}

将每个答案写成 $r\mathrm{e}^{\mathrm{i}\theta}$ 的形式，其中 $0<\theta<2\pi$ 。

(4)

解答

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把 $32\mathrm{i}$ 写成 exponential form，再取五次方根。因为五次方根在 Argand diagram 上平均分布，所以 argument 每次相差

\begin{align*} \frac{2\pi}{5}. \end{align*}

题目要求 $0<\theta<2\pi$ ，所以最后要列出这个范围内的五个角。

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From

\begin{align*} z^5-32\mathrm{i}=0, \end{align*}

we have

\begin{align*} z^5=32\mathrm{i}. \end{align*}

Now

\begin{align*} 32\mathrm{i}=32\mathrm{e}^{\frac{\pi}{2}\mathrm{i}}. \end{align*}

Let

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

Then

\begin{align*} z^5=r^5\mathrm{e}^{5\mathrm{i}\theta}. \end{align*}

\begin{align*} r^5=32, \end{align*}

giving

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

For the arguments,

\begin{align*} 5\theta=\frac{\pi}{2}+2k\pi, \end{align*}

where $k$ is an integer. Hence

\begin{align*} \theta =&\,\frac{\pi}{10}+\frac{2k\pi}{5}. \end{align*}

The values in the range $0<\theta<2\pi$ are

\begin{align*} \theta=&\,\frac{\pi}{10},\quad\frac{\pi}{2},\\[4mm] &\,\hspace{2pt}\frac{9\pi}{10},\quad\frac{13\pi}{10},\\[4mm] &\,\hspace{4pt}\frac{17\pi}{10}. \end{align*}

Therefore the five roots are

\begin{align*} z_1=&\,2\mathrm{e}^{\frac{\pi}{10}\mathrm{i}},\\[4mm] z_2=&\,2\mathrm{e}^{\frac{\pi}{2}\mathrm{i}},\\[4mm] z_3=&\,2\mathrm{e}^{\frac{9\pi}{10}\mathrm{i}},\\[4mm] z_4=&\,2\mathrm{e}^{\frac{13\pi}{10}\mathrm{i}},\\[4mm] z_5=&\,2\mathrm{e}^{\frac{17\pi}{10}\mathrm{i}}. \end{align*}