<s>[INST] 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? 
 [/INST] 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 $2^{-(2011+1/2)}(1-i)$.
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