求$$$\sqrt[3]{i}$$$
您的輸入
求$$$\sqrt[3]{i}$$$。
解答
$$$i$$$ 的極座標形式為 $$$\cos{\left(\frac{\pi}{2} \right)} + i \sin{\left(\frac{\pi}{2} \right)}$$$(步驟請參見 極座標形式計算器)。
根據棣莫弗公式,複數 $$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$ 的所有第 $$$n$$$ 次方根由 $$$r^{\frac{1}{n}} \left(\cos{\left(\frac{\theta + 2 \pi k}{n} \right)} + i \sin{\left(\frac{\theta + 2 \pi k}{n} \right)}\right)$$$, $$$k=\overline{0..n-1}$$$ 給出。
我們有 $$$r = 1$$$、$$$\theta = \frac{\pi}{2}$$$ 和 $$$n = 3$$$。
- $$$k = 0$$$: $$$\sqrt[3]{1} \left(\cos{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 0}{3} \right)} + i \sin{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 0}{3} \right)}\right) = \cos{\left(\frac{\pi}{6} \right)} + i \sin{\left(\frac{\pi}{6} \right)} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$$
- $$$k = 1$$$: $$$\sqrt[3]{1} \left(\cos{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 1}{3} \right)} + i \sin{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 1}{3} \right)}\right) = \cos{\left(\frac{5 \pi}{6} \right)} + i \sin{\left(\frac{5 \pi}{6} \right)} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$$
- $$$k = 2$$$: $$$\sqrt[3]{1} \left(\cos{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 2}{3} \right)} + i \sin{\left(\frac{\frac{\pi}{2} + 2\cdot \pi\cdot 2}{3} \right)}\right) = \cos{\left(\frac{3 \pi}{2} \right)} + i \sin{\left(\frac{3 \pi}{2} \right)} = - i$$$
答案
$$$\sqrt[3]{i} = \frac{\sqrt{3}}{2} + \frac{i}{2}\approx 0.866025403784439 + 0.5 i$$$A
$$$\sqrt[3]{i} = - \frac{\sqrt{3}}{2} + \frac{i}{2}\approx -0.866025403784439 + 0.5 i$$$A
$$$\sqrt[3]{i} = - i$$$A
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