$(1+i)^6$ 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!
 A: 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: $$(1+i)^6 = 8\left(\cos \frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right) = 8e^{3i\pi/2}$$
A: $1+i = \sqrt{2}\left(\cos (\frac{\pi}{4}) + i\sin (\frac{\pi}{4})\right)= \sqrt{2}e^{i\frac{\pi}{4}} \Rightarrow (1+i)^6 = (\sqrt{2})^6\cdot e^{i(3\frac{\pi}{2})}=-8i$
A: Hint (you made an arithmetic error):
$$ (1+i)^6 \neq \left(\frac{\sqrt2}{2}\right)^6\left(\frac{\sqrt2}{2} +\frac{\sqrt2}{2}i\right)^6$$
A: $$
\left(\frac{\sqrt 2}2\right)^6 = \left(\frac{\sqrt 2}{\sqrt 2\sqrt 2}\right)^6 = \left( \frac 1 {\sqrt 2} \right)^6 = \frac 1 {\sqrt{2}^6} =\frac 1 {(\sqrt 2^2)^3} = \frac 1 {2^3} = \frac 1 8.
$$
(But you have $\dfrac 1{4\pi}$ where you need $\dfrac 1 4 \pi$.)
A: $$(1+i)^6=\left(\sqrt{2}e^{\frac{\pi}{4}i}\right)^6=\left(\sqrt{2}\right)^6e^{6\left(\frac{\pi}{4}i\right)}=8e^{\frac{3}{2}\pi i}=8e^{-\frac{\pi}{2}i}=-8i$$
