<s>[INST] $(1+i)^6$ 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!
 [/INST] $1+i = \sqrt{2}\left(\cos (\frac{\pi}{4}) + i\sin (\frac{\pi}{4})\right)= \sqrt{2}e^{i\frac{\pi}{4}} \Rightarrow (1+i)^6 = (\sqrt{2})^6\cdot e^{i(3\frac{\pi}{2})}=-8i$
</s>