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

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

Bentuk polar dari $$$-1$$$ adalah $$$\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}$$$ (untuk langkah-langkah, lihat kalkulator bentuk polar).

Menurut Rumus De Moivre, semua akar ke-$$$n$$$ dari bilangan kompleks $$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$ diberikan oleh $$$r^{\frac{1}{n}} \left(\cos{\left(\frac{\theta + 2 \pi k}{n} \right)} + i \sin{\left(\frac{\theta + 2 \pi k}{n} \right)}\right)$$$, $$$k=\overline{0..n-1}$$$.

Diketahui bahwa $$$r = 1$$$, $$$\theta = \pi$$$, dan $$$n = 3$$$.

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

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

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

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