Halla $$$\sqrt[3]{-1}$$$

Halla $$$\sqrt[3]{-1}$$$.

La forma polar de $$$-1$$$ es $$$\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}$$$ (para ver los pasos, consulte calculadora de forma polar).

Según la fórmula de De Moivre, todas las raíces $$$n$$$-ésimas de un número complejo $$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$ vienen dadas por $$$r^{\frac{1}{n}} \left(\cos{\left(\frac{\theta + 2 \pi k}{n} \right)} + i \sin{\left(\frac{\theta + 2 \pi k}{n} \right)}\right)$$$, $$$k=\overline{0..n-1}$$$.

Tenemos que $$$r = 1$$$, $$$\theta = \pi$$$ y $$$n = 3$$$.

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

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

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

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