Evaluate $(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$ Evaluate $$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$$
So ... 
$$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011} = (-\sqrt{2}+\sqrt{2}i)^{-2011}$$
$$\theta=\pi - \arctan(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{3\pi}{4}$$
$$-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$$
$$(-\sqrt{2}+\sqrt{2}i)^{-2011}
=\cos{(-2011\theta)} + i \sin{(-2011\theta)} = e^{i(-2011)\theta}$$
Is it correct? 

The given answer is 
... 
$$\arg{z} = \frac{3\pi}{4}, \qquad z=2e^{i \frac{3\pi}{4}}$$
$$LHS 
= (\frac{1}{z})^{2011}=2^{-2011}e^{-2011(\frac{3\pi}{4})i}
= 2^{-2011} e^{-(\color{red}{1508\pi i} + \frac{\pi i }{4})}
= 2^{-2011}e^{-\frac {\pi i}{4}}
= 2^{-2011} \color{blue}{\frac{1 - i}{\sqrt{2}}}
= ...
$$
Why is the red $1508\pi i$ removed in the following step? 
How do I get the blue $\frac{1 - i}{\sqrt{2}}$ from the prev step? 
 A: First off, let's get the signs correct:
$$\left(\frac{1}{-\sqrt{2}+\sqrt{2}~i}\right)^{\color{Red}+2011}=(-\sqrt{2}+\sqrt{2}~i)^{-2011}.$$
Second, if $z=-\sqrt{2}+\sqrt{2}i=\sqrt{2}(-1+i)$, then we do have a magnitude of two:
$$|z|=\sqrt{2}\cdot |-1+i|=\sqrt{2}\cdot\sqrt{(-1)^2+1}=2.$$
The angle $\theta=\frac{3}{4}\pi$ of $z$ is correct so finally
$$z^{-2011}=(2e^{\frac{3}{4}\pi i})^{-2011}=2^{-2011}\exp\left(-\frac{6033}{4}\pi i\right)=2^{2011}e^{-\pi i/4}$$
because $6033\equiv1\bmod8$. We could also evaluate $e^{-\pi i/4}$ as $(1-i)/\sqrt{2}$ and get $2^{-(2011+1/2)}(1-i)$.
A: Ok, lots of bits here and there.
With respect to your answer, it is not true that $-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$, as in particular the length of the element on the left is $2$ and the length of the element on the right is $1$.
With respect to your question about the red bit, recall that $e^{2 i \pi} = 1$. So one can simply ignore the part of the exponent properly divisible by 2.
Similarly, they get the blue part by simply evaluating $e^{-i\pi/4 }$.
A: $$
\frac{1}{-\sqrt{2}+i\sqrt{2}}\frac{-\sqrt{2}-i\sqrt{2}}{-\sqrt{2}-i\sqrt{2}}=\frac{-\sqrt{2}-i\sqrt{2}}{4}=\frac12\frac{-1-i}{\sqrt{2}}
$$
which has absolute value $\dfrac12$ and argument $\dfrac{5\pi}{4}$, so
$\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{2011}$ has absolute value $1/2^{2011}$ and argument $\frac{5\pi}{4}\cdot2011\equiv-\frac{\pi}{4}\pmod{2\pi}$. Therefore,
$$
\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{2011}=\frac{1}{2^{2011}}\frac{1-i}{\sqrt{2}}
$$
If, as the title suggests, the exponent is $-2011$, we simply need to take the reciprocal:
$$
\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{-2011}=2^{2011}\frac{1+i}{\sqrt{2}}
$$
A: Since everyone has been answering about the blue part, I will answer about the red part:
\begin{eqnarray*} 2^{-2011} e^{-(\color{red}{1508\pi i} + \frac{\pi i }{4})} 
&=& 2^{-2011}e^{-(\color{red}{1508\pi i})} \cdot e^{\left( \frac{\pi i }{4}\right)} \\
&=& 2^{-2011}{\left(e^{(\color{red}{2\pi i})}\right)}^{-\frac{1508}{2}} \cdot e^{\left( \frac{\pi i }{4}\right)} \\
&=&  2^{-2011}{(\color{red}{1})}^{-\frac{1508}{2}} \cdot e^{\left( \frac{\pi i }{4}\right)} \hspace{5mm} \text{(Using the fact that $e^{2 \pi i } = 1$)} \\
&=& 2^{-2011} e^{\left( \frac{\pi i }{4}\right)}.  \end{eqnarray*}
