题目

Problem

Solve the equation

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

Give your answers in the form $r(\cos\theta+\mathrm{i}\sin\theta)$ , where $r>0$ and $0\leqslant\theta<2\pi$

(5)

题目中文翻译

解方程

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

将答案写成 $r(\cos\theta+\mathrm{i}\sin\theta)$ 的形式，其中 $r>0$ 且 $0\leqslant\theta<2\pi$

(5)

解答

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把 $32$ 写成复数的三角形式。因为 $32=32(\cos 0+\mathrm{i}\sin0)$ ，五次根的模长是 $2$ ，辐角要把一整圈平均分成 $5$ 份。

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Write

\begin{align*} 32=32(\cos 0+\mathrm{i}\sin0) \end{align*}

The fifth roots have modulus

\begin{align*} r=32^{1/5}=2 \end{align*}

The arguments are

\begin{align*} \theta =&\, \frac{0+2k\pi}{5}, \qquad k=0,1,2,3,4 \end{align*}

Therefore

\begin{align*} \theta =&\, 0,\frac{2\pi}{5},\frac{4\pi}{5}, \frac{6\pi}{5},\frac{8\pi}{5} \end{align*}

So the solutions are

\begin{align*} z_1 =&\, 2(\cos0+\mathrm{i}\sin0) = 2\\[3mm] z_2 =&\, 2\Big(\cos\frac{2\pi}{5}+\mathrm{i}\sin\frac{2\pi}{5}\Big)\\[3mm] z_3 =&\, 2\Big(\cos\frac{4\pi}{5}+\mathrm{i}\sin\frac{4\pi}{5}\Big)\\[3mm] z_4 =&\, 2\Big(\cos\frac{6\pi}{5}+\mathrm{i}\sin\frac{6\pi}{5}\Big)\\[3mm] z_5 =&\, 2\Big(\cos\frac{8\pi}{5}+\mathrm{i}\sin\frac{8\pi}{5}\Big) \end{align*}