### Instruction: Compute $(-1 + i \sqrt{3})^8.$  Enter the answer in rectangular form.### Response: I know that I can use the polar form of complex numbers to simplify this computation.
I can write $-1 + i \sqrt{3}$ as $2 \text{cis}(\frac{2\pi}{3})$, where $\text{cis}(\theta) = \cos \theta + i \sin \theta.$
Then I can use the formula $(r \text{cis}(\theta))^n = r^n \text{cis}(n \theta)$ to raise this to the eighth power.
So I have $(-1 + i \sqrt{3})^8 = (2 \text{cis}(\frac{2\pi}{3}))^8 = 2^8 \text{cis}(\frac{16 \pi}{3}).$
I can simplify this by noting that $\text{cis}(\frac{16 \pi}{3}) = \text{cis}(\frac{4 \pi}{3}),$ since adding a multiple of $2 \pi$ to the angle does not change the complex number.
So I have $(-1 + i \sqrt{3})^8 = 256 \text{cis}(\frac{4 \pi}{3}).$
To convert this back to rectangular form, I need to use the formula $r \text{cis}(\theta) = r \cos \theta + i r \sin \theta.$
So I have $(-1 + i \sqrt{3})^8 = 256 \cos \frac{4 \pi}{3} + i 256 \sin \frac{4 \pi}{3}.$
I can use the fact that $\cos \frac{4 \pi}{3} = -\frac{1}{2}$ and $\sin \frac{4 \pi}{3} = -\frac{\sqrt{3}}{2}$ to simplify this further.
So I have $(-1 + i \sqrt{3})^8 = -128 - i 128 \sqrt{3}.$