## Precalculus (6th Edition) Blitzer

$1$, $-\frac{1}{2}+\frac{\sqrt 3}{2}i$, $-\frac{1}{2}-\frac{\sqrt 3}{2}i$
Step 1. Let $z=1$; we have $z=cos0+i\ sin0$ Step 2. Based on De Moivre's Theorem, we have the cube roots as $z_k=\sqrt[3] 1(cos\frac{2k\pi+0}{3}+i\ sin\frac{2k\pi+0}{3})$ where $k=0,1,2$ Step 3. For $k=0$, we have $z_0=1$ Step 4. For $k=1$, we have $z_1=cos\frac{2\pi}{3}+i\ sin\frac{2\pi}{3}=-\frac{1}{2}+\frac{\sqrt 3}{2}i$ Step 5. For $k=2$, we have $z_2=cos\frac{4\pi}{3}+i\ sin\frac{4\pi}{3}=-\frac{1}{2}-\frac{\sqrt 3}{2}i$