$(1+i)^6$ 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!

Hint (you made an arithmetic error):
$$ (1+i)^6 \neq \left(\frac{\sqrt2}{2}\right)^6\left(\frac{\sqrt2}{2} +\frac{\sqrt2}{2}i\right)^6$$
