Find all the roots of the complex number \(z = 8(cos(\frac{\pi}{3}) + i sin(\frac{\pi}{3}))\).

Step 1 ：Step 1: We write it in exponential form \(z = 8e^{\frac{i\pi}{3}}\)

Step 2 ：Step 2: We know that the roots of the complex number \(z^n = r^n(cos(n\theta) + i sin(n\theta))\) are given by \(z_k = r^{\frac{1}{n}}(cos(\frac{\theta + 2k\pi}{n}) + i sin(\frac{\theta + 2k\pi}{n}))\) where \(k = 0, 1, 2, \ldots, n - 1\).

Step 3 ：Step 3: We need to find the roots of the equation \(z^n = 8e^{\frac{i\pi}{3}}\).

Step 4 ：Step 4: For \(n = 1\), the root is \(z_0 = 8^{\frac{1}{1}}(cos(\frac{\frac{\pi}{3} + 2 * 0 * \pi}{1}) + i sin(\frac{\frac{\pi}{3} + 2 * 0 * \pi}{1})) = 8(cos(\frac{\pi}{3}) + i sin(\frac{\pi}{3})) = 8e^{\frac{i\pi}{3}}\).

Step 5 ：Step 5: For \(n = 2\), the roots are \(z_0 = 8^{\frac{1}{2}}(cos(\frac{\frac{\pi}{3} + 2 * 0 * \pi}{2}) + i sin(\frac{\frac{\pi}{3} + 2 * 0 * \pi}{2})) = 2\sqrt{2}(cos(\frac{\pi}{6}) + i sin(\frac{\pi}{6})) = 2\sqrt{2}e^{\frac{i\pi}{6}}\) and \(z_1 = 8^{\frac{1}{2}}(cos(\frac{\frac{\pi}{3} + 2 * 1 * \pi}{2}) + i sin(\frac{\frac{\pi}{3} + 2 * 1 * \pi}{2})) = 2\sqrt{2}(cos(\frac{5\pi}{6}) + i sin(\frac{5\pi}{6})) = 2\sqrt{2}e^{\frac{5i\pi}{6}}\).

Step 6 ：Step 6: For \(n = 3\), the roots are \(z_0 = 8^{\frac{1}{3}}(cos(\frac{\frac{\pi}{3} + 2 * 0 * \pi}{3}) + i sin(\frac{\frac{\pi}{3} + 2 * 0 * \pi}{3})) = 2(cos(\frac{\pi}{9}) + i sin(\frac{\pi}{9})) = 2e^{\frac{i\pi}{9}}\), \(z_1 = 8^{\frac{1}{3}}(cos(\frac{\frac{\pi}{3} + 2 * 1 * \pi}{3}) + i sin(\frac{\frac{\pi}{3} + 2 * 1 * \pi}{3})) = 2(cos(\frac{7\pi}{9}) + i sin(\frac{7\pi}{9})) = 2e^{\frac{7i\pi}{9}}\), and \(z_2 = 8^{\frac{1}{3}}(cos(\frac{\frac{\pi}{3} + 2 * 2 * \pi}{3}) + i sin(\frac{\frac{\pi}{3} + 2 * 2 * \pi}{3})) = 2(cos(\frac{13\pi}{9}) + i sin(\frac{13\pi}{9})) = 2e^{\frac{13i\pi}{9}}\).