Questions: The complex number (-1-sqrt(3) i)^8 is equal to Select one: 2^7 e^(i pi/3) 2^7+2^7 sqrt(3) i -2^7-2^7 sqrt(3) i 2^8 e^(i 2/3 pi) -2^8+2^8 sqrt(3) i

Solution Steps

To convert the complex number \(-1 - \sqrt{3} i\) to polar form, we first calculate the modulus \(r\) and the argument \(\theta\). The modulus is given by: \[ r = |z| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \] The argument \(\theta\) can be found using: \[ \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{-1}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \] Since the complex number is in the third quadrant, we adjust \(\theta\) to: \[ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \] Thus, the polar form is: \[ -1 - \sqrt{3} i = 2 e^{i \frac{4\pi}{3}} \]

Using De Moivre's Theorem, we raise the polar form to the 8th power: \[ (-1 - \sqrt{3} i)^{8} = (2 e^{i \frac{4\pi}{3}})^{8} = 2^8 e^{i \frac{32\pi}{3}} \] Next, we simplify the exponent: \[ \frac{32\pi}{3} = 10\pi + \frac{2\pi}{3} \quad \text{(since } 10\pi \text{ is a multiple of } 2\pi\text{)} \] Thus, we can reduce it to: \[ e^{i \frac{2\pi}{3}} \]

Combining the results, we have: \[ (-1 - \sqrt{3} i)^{8} = 2^8 e^{i \frac{2\pi}{3}} \] This matches the form of the options provided in the original question.

Final Answer