IAL 2025 June Q1 | 國際數學站

(a) Express $-2\sqrt{2} - (2\sqrt{6})\mathrm{i}$ in the form $re^{\mathrm{i}\theta}$ where $-\pi < \theta \leqslant \pi$

(3)

(b) Hence solve

\begin{align*} z^5 = -2\sqrt{2} - (2\sqrt{6})\mathrm{i} \end{align*}

Give your answers in the form $\sqrt{p}e^{\mathrm{i}\theta}$ where $p \in \mathbb{Z}^{+}$ and $-\pi < \theta \leqslant \pi$

(3)

解答

(a)

For $z = -2\sqrt{2} - 2\sqrt{6}\mathrm{i}$ , the modulus $r$ is:

\begin{align*} r = &\,\sqrt{(-2\sqrt{2})^2 + (-2\sqrt{6})^2}\\[4mm] = &\,\sqrt{8 + 24}\\[4mm] = &\,\sqrt{32} = 4\sqrt{2} \end{align*}

The argument $\theta$ is in the third quadrant:

\begin{align*} \tan\alpha = &\,\frac{2\sqrt{6}}{2\sqrt{2}} = \sqrt{3} \implies \alpha = \frac{\pi}{3}\\[4mm] \theta = &\,-\pi + \frac{\pi}{3} = -\frac{2\pi}{3} \end{align*}

Thus, in the form $re^{\mathrm{i}\theta}$ :

\begin{align*} -2\sqrt{2} - 2\sqrt{6}\mathrm{i} = &\,4\sqrt{2}\mathrm{e}^{-\frac{2\pi}{3}\mathrm{i}} \end{align*}

(b)

\begin{align*} z^5 = &\,32^{\frac{1}{2}}\mathrm{e}^{\mathrm{i}\left(-\frac{2\pi}{3} + 2k\pi\right)}\\[4mm] z = &\,(32^{\frac{1}{2}})^{\frac{1}{5}}\mathrm{e}^{\mathrm{i}\left(\frac{-\frac{2\pi}{3} + 2k\pi}{5}\right)}\\[4mm] = &\,\sqrt{2}\mathrm{e}^{\mathrm{i}\left(-\frac{2\pi}{15} + \frac{6k\pi}{15}\right)} \qquad k = -2, -1, 0, 1, 2 \end{align*}

By substituting the values of $k$ , we obtain the five roots (ensuring $-\pi < \theta \leqslant \pi$ ):

\begin{align*} k = 0 \implies &\, z_1 = \sqrt{2}\mathrm{e}^{-\frac{2\pi}{15}\mathrm{i}}\\[4mm] k = 1 \implies &\, z_2 = \sqrt{2}\mathrm{e}^{\frac{4\pi}{15}\mathrm{i}}\\[4mm] k = 2 \implies &\, z_3 = \sqrt{2}\mathrm{e}^{\frac{10\pi}{15}\mathrm{i}} = \sqrt{2}\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}\\[4mm] k = -1 \implies &\, z_4 = \sqrt{2}\mathrm{e}^{-\frac{8\pi}{15}\mathrm{i}}\\[4mm] k = -2 \implies &\, z_5 = \sqrt{2}\mathrm{e}^{-\frac{14\pi}{15}\mathrm{i}} \end{align*}