Hello Matt,
What we want to do is the get the complex values to have coefficients that represent the cosine (for the real part) and the sine (for the imaginary part) of some common angle, so that we can then apply de Moivre's theorem.
1.) $$z=\left(\sqrt{3}+i \right)^5$$
Now, if we factor out $\dfrac{1}{2}$ from the complex value, we may write:
$$z=2^5\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right)^5$$
Now since:
$$\cos\left(\frac{\pi}{6} \right)=\frac{\sqrt{3}}{2},\,\sin\left(\frac{\pi}{6} \right)=\frac{1}{2}$$
we obtain:
$$z=2^5\left(\cos\left(\frac{\pi}{6} \right)+i\sin\left(\frac{\pi}{6} \right) \right)^5$$
Applying de Moivre's theorem, we have:
$$z=2^5\left(\cos\left(\frac{5\pi}{6} \right)+i\sin\left(5\frac{\pi}{6} \right) \right)=32\left(-\frac{\sqrt{3}}{2}+\frac{1}{2} \right)=-16\sqrt{3}+16i$$
2.) $$z=(2-2\sqrt{3}i)^4$$
Factoring out $$4$$ we may write:
$$z=4^4\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i \right)^4$$
Replacing the coefficients with trigonometric functions, we have:
$$z=4^4\left(\cos\left(-\frac{\pi}{3} \right)+i\sin\left(-\frac{\pi}{3} \right) \right)^4$$
Applying de Moivre's theorem, we get:
$$z=4^4\left(\cos\left(-\frac{4\pi}{3} \right)+i\sin\left(-\frac{4\pi}{3} \right) \right))=256\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i \right)=-128+128\sqrt{3}i$$
3.) $$z=(-2-2i)^5$$
Factoring out $$-2\sqrt{2}$$ we get:
$$z=-2^{\frac{15}{2}}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i \right)^5$$
Replacing the coefficients with trigonometric functions, we have:
$$z=-2^{\frac{15}{2}}\left(\cos\left(\frac{\pi}{4} \right)+i\sin\left(\frac{\pi}{4} \right) \right)^5$$
Applying de Moivre's theorem, we get:
$$z=-2^{\frac{15}{2}}\left(\cos\left(\frac{5\pi}{4} \right)+i\sin\left(\frac{5\pi}{4} \right) \right)=-2^{\frac{15}{2}}\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i \right)=128+128i$$