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? 

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*}
