TITLE: Find the 2016th power of a complex number
QUESTION [8 upvotes]: 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!

REPLY [2 votes]: The main trick that makes this easy is to immediately recognize by sight parts of the third and eighth roots of unity.
The two primitive third roots of unity are
$$ \frac{-1 \pm \mathbf{i} \sqrt{3}}{2} $$
and the four primitive eighth roots of unity are
$$ \frac{\pm 1 \pm \mathbf{i}}{\sqrt{2}} $$
Sometimes people like to write $\frac{\sqrt{2}}{2}$ rather than $\frac{1}{\sqrt{2}}$.
And since they're similar, I'll mention the two primitive sixth roots of unity are
$$ \frac{1 \pm \mathbf{i} \sqrt{3}}{2} $$
In each case, the root with smallest positive complex argument is the one where you take the positive signs.
With this in mind, we rewrite the base as
$$\frac{-1 + i\sqrt 3}{1 + i} 
= \frac{-1 + \mathbf{i} \sqrt{3}}{2} \frac{\sqrt{2}}{1 + \mathbf{i}} \sqrt{2} 
= \sqrt{2} \zeta_3 \zeta_8^{-1} $$
where I've used $\zeta_n$ to denote the principal $n$-th root of unity.
Now, the 2016-th power is easy to compute!