<s>[INST] Find the 2016th power of a complex number Calculate $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$.
Here is what I did so far:
I'm trying to transform $z$ into its trigonometric form, so I can use De Moivre's formula for calculating $z^{2016}$.
Let $z = \frac{-1 + i\sqrt 3}{1 + i}$. This can be rewritten as $\frac{\sqrt 3 - 1}{2} + i\frac{\sqrt 3 + 1}{2}$.
$$z = r(\cos \phi + i \sin \phi)$$
$$r = |z| = \sqrt 2$$
$$\phi = \arctan {\sqrt 3 + 1}$$
Now, I don't know what to do with that $\sqrt 3 + 1$. How do I calculate $\phi$ ?
Thank you in advance!
 [/INST] $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$
Lets simplify $\frac{-1 + i\sqrt 3}{1 + i}$
$\frac {1}{1+i}(-1 + i\sqrt 3)\\
\frac {1-i}{2}(-1 + i\sqrt 3)\\
 (1-i)(-\frac12 + i\frac{\sqrt 3}2)$
Convert to polar:
$\sqrt 2 (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3)$
Now we have a choice...we could raise to the 2016 power right now, or we could mulitiply those two complex numbers first then raise to the 2016 power.
$\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016} =
\sqrt 2^{2016} (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016}(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3)^{2016}$
Applying deMoivres theorem:
$(\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016} = (\cos \frac {-2016\pi}{4} +\sin \frac {-2016\pi}{4} )$
$8$ divides $2016$ 
$\frac {-2016\pi}{4} = 2n\pi$
$(\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016} = 1$
$6$ divides $2016$, too.
$2^{1008}$
alternatively:
$\sqrt 2 (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3) = \sqrt 2 (\cos \frac {5\pi}{12} +\sin \frac {5\pi}{12})$
and we get to the same place.
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