$$$\sqrt[3]{-8}$$$을(를) 구하시오

$$$\sqrt[3]{-8}$$$을(를) 구하시오.

$$$-8$$$의 극형식은 $$$8 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right)$$$입니다(풀이 단계는 극형식 계산기를 참조하세요).

드무아브르의 공식에 따르면, 복소수 $$$r \left(\cos{\left(\theta \right)} + i \sin{\left(\theta \right)}\right)$$$의 모든 $$$n$$$제곱근은 $$$r^{\frac{1}{n}} \left(\cos{\left(\frac{\theta + 2 \pi k}{n} \right)} + i \sin{\left(\frac{\theta + 2 \pi k}{n} \right)}\right)$$$, $$$k=\overline{0..n-1}$$$로 주어진다.

다음이 성립한다: $$$r = 8$$$, $$$\theta = \pi$$$, 및 $$$n = 3$$$.

• $$$k = 0$$$: $$$\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 0}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 0}{3} \right)}\right) = 2 \left(\cos{\left(\frac{\pi}{3} \right)} + i \sin{\left(\frac{\pi}{3} \right)}\right) = 1 + \sqrt{3} i$$$
• $$$k = 1$$$: $$$\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 1}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 1}{3} \right)}\right) = 2 \left(\cos{\left(\pi \right)} + i \sin{\left(\pi \right)}\right) = -2$$$
• $$$k = 2$$$: $$$\sqrt[3]{8} \left(\cos{\left(\frac{\pi + 2\cdot \pi\cdot 2}{3} \right)} + i \sin{\left(\frac{\pi + 2\cdot \pi\cdot 2}{3} \right)}\right) = 2 \left(\cos{\left(\frac{5 \pi}{3} \right)} + i \sin{\left(\frac{5 \pi}{3} \right)}\right) = 1 - \sqrt{3} i$$$

$$$\sqrt[3]{-8} = 1 + \sqrt{3} i\approx 1 + 1.732050807568877 i$$$A

$$$\sqrt[3]{-8} = -2$$$A

$$$\sqrt[3]{-8} = 1 - \sqrt{3} i\approx 1 - 1.732050807568877 i$$$A