求$$$\sqrt[3]{-1}$$$

求$$$\sqrt[3]{-1}$$$。

$$$-1$$$ 的极坐标形式是 $$$\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}$$$（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$$$r = 1$$$、$$$\theta = \pi$$$和$$$n = 3$$$。

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

$$$\sqrt[3]{-1} = \frac{1}{2} + \frac{\sqrt{3} i}{2}\approx 0.5 + 0.866025403784439 i$$$A

$$$\sqrt[3]{-1} = -1$$$A

$$$\sqrt[3]{-1} = \frac{1}{2} - \frac{\sqrt{3} i}{2}\approx 0.5 - 0.866025403784439 i$$$A