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?