<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] $$
\left(\frac{\sqrt 2}2\right)^6 = \left(\frac{\sqrt 2}{\sqrt 2\sqrt 2}\right)^6 = \left( \frac 1 {\sqrt 2} \right)^6 = \frac 1 {\sqrt{2}^6} =\frac 1 {(\sqrt 2^2)^3} = \frac 1 {2^3} = \frac 1 8.
$$
(But you have $\dfrac 1{4\pi}$ where you need $\dfrac 1 4 \pi$.)
</s>