题目

Problem

(a) Express the complex number $18\sqrt3-18\mathrm{i}$ in the form

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

(3)

(b) Solve the equation

\begin{align*} z^4=18\sqrt3-18\mathrm{i} \end{align*}

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

(5)

解答

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先求模长，再找辐角。实部为正、虚部为负，所以复数在第四象限，辐角应为负角。

#### 答题过程

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The modulus is

\begin{align*} r =&\,\sqrt{(18\sqrt3)^2+(-18)^2}\\[4mm] =&\,\sqrt{972+324}\\[4mm] =&\,36 \end{align*}

For the argument,

\begin{align*} \tan\theta =&\,\frac{-18}{18\sqrt3}\\[4mm] =&\,-\frac1{\sqrt3} \end{align*}

Since the complex number is in the fourth quadrant,

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

Therefore

\begin{align*} 18\sqrt3-18\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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四次根的模长是 $36^{1/4}=\sqrt6$ 。辐角要把 $-\dfrac{\pi}{6}+2k\pi$ 除以 $4$ ，取四个互不相同且在指定范围内的角。

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From part (a),

\begin{align*} z^4 =&\,36\left(\cos\left(-\frac{\pi}{6}\right) +\mathrm{i}\sin\left(-\frac{\pi}{6}\right)\right) \end{align*}

Thus

\begin{align*} z =&\,\sqrt6 \left[ \cos\left(\frac{-\frac{\pi}{6}+2k\pi}{4}\right) +\mathrm{i}\sin\left(\frac{-\frac{\pi}{6}+2k\pi}{4}\right) \right]\\[4mm] =&\,\sqrt6 \left[ \cos\left(-\frac{\pi}{24}+\frac{k\pi}{2}\right) +\mathrm{i}\sin\left(-\frac{\pi}{24}+\frac{k\pi}{2}\right) \right] \end{align*}

for $k=0,1,2,3$ . The arguments are

\begin{align*} -\frac{\pi}{24},\quad \frac{11\pi}{24},\quad \frac{23\pi}{24},\quad \frac{35\pi}{24} \end{align*}

The last angle is greater than $\pi$ , so subtract $2\pi$ :

\begin{align*} \frac{35\pi}{24}-2\pi=-\frac{13\pi}{24} \end{align*}

Therefore the roots are

\begin{align*} \sqrt6e^{-\frac{\pi}{24}\mathrm{i}},\quad \sqrt6e^{\frac{11\pi}{24}\mathrm{i}},\quad \sqrt6e^{\frac{23\pi}{24}\mathrm{i}},\quad \sqrt6e^{-\frac{13\pi}{24}\mathrm{i}} \end{align*}