$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got 
$$
  \left(\frac{\sqrt{2}}{2}\right)^6 \times 
  \cos\left(6 \times \frac{1}{4\pi}\right) + 
  i\sin\left(6 \times \frac{1}{4\pi}\right) = 
  \frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where the $8$ is from? 
Thanks!

We have that $|1+i| = \sqrt{2}$ and ${\rm Arg}(1+i) = \pi/4$. So: $$1+i = \sqrt{2}\left(\cos \frac{\pi}{4}+i \sin\frac{\pi}{4}\right).$$By De Moivre's formula: $$(1+i)^6 = (\sqrt{2})^6 \left(\cos \frac{6\pi}{4}+i\sin\frac{6\pi}{4}\right).$$But $\sqrt{2}^6 = (2^{1/2})^6 = 2^3=8$. Simplifying: $$(1+i)^6 = 8\left(\cos \frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right) = 8e^{3i\pi/2}$$
