complex number with high powers The question:
$\frac{(-1-\sqrt{3}i)^{73}}{2^{73}}$
I really do not even know where to begin. Am I suppose to expand $(-1-\sqrt{3}i)$ 73 times? 
 A: Hint:
$$\frac{(-1-\sqrt{3}i)^{73}}{2^{73}}=(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)^{73}$$
and
$$(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)^{3}=1$$
A: Hint:
$$\frac{-1-\sqrt{3} i}{2} = r e^{i \theta}$$
for some real number $\theta$ and nonnegative number $r$. Can you find what $\theta$ and $r$ are? Euler's formula may be helpful. Raising $r e^{i \theta}$ to the $73$rd power is a bit easier.
A: Hint: $\quad\dfrac{-1-i\sqrt 3}2=\mathrm e^{\tfrac{4i\pi}3}.$
A: A systematic approach goes like this.
1) Represent any complex number $z\in\mathbb{C}$, your example being $z=\frac{-1-\sqrt{3} i}{2}$ in polar coordinates
$$z = r e^{i \theta},$$
where $$r=\sqrt{\textrm{Re}{z}^2+\textrm{Im}{z}^2}$$ and $$\theta = \arg{z} = \arctan{\frac{\textrm{Re}{z}}{\textrm{Im}{z}}}$$ unless $\textrm{Im}{z}=0$.
In your example, we find $r=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$ and $\theta=-\frac{2}{3}2\pi$
2) Take the power $$z^n=r^ne^{i\theta n}$$ for $n\in\mathbb{Z}$. In your example, we obtain
$z^{73}=1^{73}e^{-i73\frac{2}{3}2\pi}=e^{-i\frac{2}{3}2\pi}=z$, because $e^{-i72\frac{2}{3}2\pi}=e^{-i48\times 2\pi}=1$.
A: Ok, what I got was:
$(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)^{73}$
$|z| = \sqrt{(-1/2)^2+\sqrt{3}/2} = \sqrt{1} = 1$ 
tan$^{-1}$ ($-.5/ \sqrt{3}/2) = \frac{-2}{3}2\pi$
$z^{73} = 1($cos$(\frac{-2}{3}2\pi)+i$sin$(\frac{-2}{3}2\pi)$) 
$z = 1^{73}($cos$(\frac{-2}{3}2\pi*73)+i$sin$(\frac{-2}{3}2\pi8*73)$) 
= $-.5 + \frac{\sqrt{3}}{2}$ which is the same answer had I not multiplied by 73
