IAL 2020 Oct Q4 | 國際數學站

(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 \end{align*}

(3)

(b) Solve the equation

\begin{align*} z^4 = 18\sqrt{3} - 18\mathrm{i} \end{align*}

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

(5)

解答

(a)

\begin{align*} 18\sqrt{3} - 18\mathrm{i} &= 18(\sqrt{3} - \mathrm{i})\\[4mm] &= 18 \cdot 2 \left(\cos\left(-\frac{\pi}{6}\right) + \mathrm{i}\sin\left(-\frac{\pi}{6}\right)\right)\\[4mm] &= 36\left(\cos\left(-\frac{\pi}{6}\right) + \mathrm{i}\sin\left(-\frac{\pi}{6}\right)\right) \end{align*}

(b)

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

So the roots are

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