求$$$\sqrt[3]{i}$$$

求$$$\sqrt[3]{i}$$$。

$$$i$$$ 的极坐标形式是 $$$\cos{\left(\frac{\pi}{2} \right)} + i \sin{\left(\frac{\pi}{2} \right)}$$$（步骤请参见 极坐标形式计算器）。

根据棣莫弗公式，复数$$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$的所有$$$n$$$次方根由$$$r^{\frac{1}{n}} \left(\cos{\left(\frac{\theta + 2 \pi k}{n} \right)} + i \sin{\left(\frac{\theta + 2 \pi k}{n} \right)}\right)$$$, $$$k=\overline{0..n-1}$$$给出。

我们有$$$r = 1$$$、$$$\theta = \frac{\pi}{2}$$$和$$$n = 3$$$。

• $$$k = 0$$$: $$$\sqrt[3]{1} \left(\cos{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 0}{3} \right)} + i \sin{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 0}{3} \right)}\right) = \cos{\left(\frac{\pi}{6} \right)} + i \sin{\left(\frac{\pi}{6} \right)} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$$
• $$$k = 1$$$: $$$\sqrt[3]{1} \left(\cos{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 1}{3} \right)} + i \sin{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 1}{3} \right)}\right) = \cos{\left(\frac{5 \pi}{6} \right)} + i \sin{\left(\frac{5 \pi}{6} \right)} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$$
• $$$k = 2$$$: $$$\sqrt[3]{1} \left(\cos{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 2}{3} \right)} + i \sin{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 2}{3} \right)}\right) = \cos{\left(\frac{3 \pi}{2} \right)} + i \sin{\left(\frac{3 \pi}{2} \right)} = - i$$$

$$$\sqrt[3]{i} = \frac{\sqrt{3}}{2} + \frac{i}{2}\approx 0.866025403784439 + 0.5 i$$$A

$$$\sqrt[3]{i} = - \frac{\sqrt{3}}{2} + \frac{i}{2}\approx -0.866025403784439 + 0.5 i$$$A

$$$\sqrt[3]{i} = - i$$$A