<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] $$
\frac{1}{-\sqrt{2}+i\sqrt{2}}\frac{-\sqrt{2}-i\sqrt{2}}{-\sqrt{2}-i\sqrt{2}}=\frac{-\sqrt{2}-i\sqrt{2}}{4}=\frac12\frac{-1-i}{\sqrt{2}}
$$
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:
$$
\left(\frac{1}{-\sqrt{2}+i\sqrt{2}}\right)^{-2011}=2^{2011}\frac{1+i}{\sqrt{2}}
$$
</s>