Bepaal $$$\sqrt[3]{-8}$$$
Uw invoer
Bepaal $$$\sqrt[3]{-8}$$$.
Oplossing
De polaire vorm van $$$-8$$$ is $$$8 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right)$$$ (voor de stappen, zie rekenmachine voor polaire vorm).
Volgens de formule van De Moivre worden alle $$$n$$$-de wortels van een complex getal $$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$ gegeven door $$$r^{\frac{1}{n}} \left(\cos{\left(\frac{\theta + 2 \pi k}{n} \right)} + i \sin{\left(\frac{\theta + 2 \pi k}{n} \right)}\right)$$$, $$$k=\overline{0..n-1}$$$.
We hebben dat $$$r = 8$$$, $$$\theta = \pi$$$ en $$$n = 3$$$.
- $$$k = 0$$$: $$$\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 0}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 0}{3} \right)}\right) = 2 \left(\cos{\left(\frac{\pi}{3} \right)} + i \sin{\left(\frac{\pi}{3} \right)}\right) = 1 + \sqrt{3} i$$$
- $$$k = 1$$$: $$$\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 1}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 1}{3} \right)}\right) = 2 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right) = -2$$$
- $$$k = 2$$$: $$$\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 2}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 2}{3} \right)}\right) = 2 \left(\cos{\left(\frac{5 \pi}{3} \right)} + i \sin{\left(\frac{5 \pi}{3} \right)}\right) = 1 - \sqrt{3} i$$$
Antwoord
$$$\sqrt[3]{-8} = 1 + \sqrt{3} i\approx 1 + 1.732050807568877 i$$$A
$$$\sqrt[3]{-8} = -2$$$A
$$$\sqrt[3]{-8} = 1 - \sqrt{3} i\approx 1 - 1.732050807568877 i$$$A
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