题目

Problem

(a) Express the complex number $18\sqrt{3}-18\mathrm{i}$ in the form

\begin{align*} r(\cos \theta + \mathrm{i}\sin \theta) \qquad -\pi < \theta \leqslant \pi\\[2mm] \end{align*}

(3)

(b) Solve the equation

\begin{align*} z^4 =\,& 18\sqrt{3}-18\mathrm{i}\\[2mm] \end{align*}

giving your answers in the form $r\mathrm{e}^{\mathrm{i}\theta}$ where $-\pi < \theta \leqslant \pi$

(5)

解答

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把复数写成模长和辐角。实部为正、虚部为负，所以点在第四象限，辐角应为负角。

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For

\begin{align*} 18\sqrt{3}-18\mathrm{i}, \end{align*}

the modulus is

\begin{align*} r=&\,\sqrt{(18\sqrt{3})^2+(-18)^2}\\[2mm] =&\,\sqrt{972+324}\\[2mm] =&\,\sqrt{1296}\\[2mm] =&\,36. \end{align*}

The argument satisfies

\begin{align*} \tan\theta=\frac{-18}{18\sqrt{3}} =-\frac{1}{\sqrt{3}}. \end{align*}

Since the complex number is in the fourth quadrant,

\begin{align*} \theta=-\frac{\pi}{6}. \end{align*}

Therefore

\begin{align*} \boxed{ 18\sqrt{3}-18\mathrm{i} =36\left(\cos\left(-\frac{\pi}{6}\right) +\mathrm{i}\sin\left(-\frac{\pi}{6}\right)\right) }. \end{align*}

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由 (a)，右边的模长是 $36$ ，辐角是 $-\frac{\pi}{6}$ 。求四次根时，模长开四次方：

\begin{align*} 36^{1/4}=\sqrt{6}. \end{align*}

辐角要先加上 $2k\pi$ ，再除以 $4$ 。

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From part (a),

\begin{align*} 18\sqrt{3}-18\mathrm{i} =36\mathrm{e}^{-\frac{\pi}{6}\mathrm{i}}. \end{align*}

\begin{align*} z^4=36\mathrm{e}^{-\frac{\pi}{6}\mathrm{i}}. \end{align*}

The fourth roots are

\begin{align*} z =&\,36^{1/4} \mathrm{e}^{\mathrm{i}\left(\frac{-\frac{\pi}{6}+2k\pi}{4}\right)}\\[2mm] =&\,\sqrt{6}\, \mathrm{e}^{\mathrm{i}\left(\frac{12k\pi-\pi}{24}\right)}, \qquad k=0,1,2,3. \end{align*}

Therefore the roots are

\begin{align*} \boxed{ \sqrt{6}\mathrm{e}^{-\frac{\pi}{24}\mathrm{i}}, \quad \sqrt{6}\mathrm{e}^{\frac{11\pi}{24}\mathrm{i}}, \quad \sqrt{6}\mathrm{e}^{\frac{23\pi}{24}\mathrm{i}}, \quad \sqrt{6}\mathrm{e}^{-\frac{13\pi}{24}\mathrm{i}} }. \end{align*}