Find the three roots of \( z=8(\cos \pi+i \sin \pi) \). Express your answer in standard form.

To find the three roots of \( z=8(\cos \pi+i \sin \pi) \), we first express it in polar form. The modulus is 8 and the argument is \( \pi \). Using De Moivre’s Theorem, the \( n \)-th roots of a complex number in polar form can be calculated as follows: 1. The modulus of the roots is \( \sqrt[3]{8} = 2 \). 2. The arguments are given by \( \frac{\pi + 2k\pi}{3} \) where \( k=0, 1, 2 \). Calculating for each value of \( k \): - For \( k=0 \): \[ \theta_0 = \frac{\pi + 0}{3} = \frac{\pi}{3} \] The root is \( 2\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right) = 2\left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 1 + i\sqrt{3} \). - For \( k=1 \): \[ \theta_1 = \frac{\pi + 2\pi}{3} = \pi \] The root is \( 2\left(\cos \pi + i \sin \pi\right) = 2(-1 + 0i) = -2 + 0i \). - For \( k=2 \): \[ \theta_2 = \frac{\pi + 4\pi}{3} = \frac{5\pi}{3} \] The root is \( 2\left(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3}\right) = 2\left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = 1 - i\sqrt{3} \). Thus, the three roots of \( z=8(\cos \pi+i \sin \pi) \) are: \[ 1 + i\sqrt{3}, \quad -2, \quad 1 - i\sqrt{3}. \]