Expansion of complex equation. Find the value of $$\left(\frac{-1+\sqrt 3i}{2}\right)^{15}  + \left(\frac{-1-\sqrt 3i}{2}\right)^{15}.$$
In general, how do we find the value of expansion of equation of high orders other than binomial expansion? 
 A: Obviously the length of $a:=\frac{-1+\sqrt{3}i}{2}$ equals $1$ and its argument is $2\pi/3$.  Hence $a^{15}=1$. Analog $\left(\frac{-1-\sqrt{3}i}{2}\right)^{15}=1$.  Hence their sum equals $2$.
A: If you notice that $\left(\frac{-1\pm\sqrt 3i}{2}\right)^3=1$, the rest should be easy.

Let me suggest another possible solution, although perhaps not the quickest way.
Using the fact that the given numbers are the roots of the quadratic equation $x^2+x+1=0$, you should be able to show that
$$a_n=\left(\frac{-1+\sqrt 3i}{2}\right)^{n}  + \left(\frac{-1-\sqrt 3i}{2}\right)^{n}$$
fulfills the recurrence relation $a_{n+2}+a_{n+1}+a_n=0$. (For example, you can prove this by induction.)
Now you can find $a_{15}$ using $a_0=2$, $a_1=-1$ and $a_{n+2}=-a_{n+1}-a_{n+1}$.
If you try this, you should notice a repeating patter after a few terms.
A: In general, if you want to find powers of a complex number, write it in polar form i.e. in the form of $r e^{i \theta} $ so that $(r e^{i \theta})^n = r^n e^{i n \theta} $. Then you can convert it back to $a + ib$ form easily as $r^n \cos(n \theta) + i r^n \sin(n \theta) $.
A: Hint:$e^{i\theta}=\cos(\theta)+i\sin(\theta)$
Also you can use DeMovire's Theorem.
A: And more generally, to find an irreducible representation of the field we work on will ensure that matrix exponentiation reduces to exponentiation of the individual blocks of the representation matrix. Such a matrix for $re^{i\theta}$ is $$\left(\begin{array}{ccc} r&0&0 \\ 0&1&0\\0&\theta&1 \end{array}\right)$$ since:
$$\left(\begin{array}{ccc} r_1&0&0 \\ 0&1&0\\0&\theta_1&1 \end{array}\right) \left(\begin{array}{ccc} r_2&0&0 \\ 0&1&0\\0&\theta_2&1 \end{array}\right) = \left(\begin{array}{ccc} r_1r_2&0&0 \\ 0&1&0\\0&\theta_1+\theta_2&1 \end{array}\right)$$
Then it is trivial to show that: $$\prod_{i=1}^N \left(\begin{array}{ccc} r_i&0&0 \\ 0&1&0\\0&\theta_i&1 \end{array}\right) = \left(\begin{array}{ccc} \prod_{i=1}^N r_i&0&0 \\ 0&1&0\\0&\sum_{I=1}^N\theta_i&1 \end{array}\right)$$ just as with polar coordinates.
A: Exploiting Martin Sleziak's comment, we notice the two addends are fifth powers of one, since the bases of both those powers are cubic roots of one. Therefore that sum is $1+1=2$.
A: $$\left(\frac{-1+\sqrt 3i}{2}\right)^{15}  + \left(\frac{-1-\sqrt 3i}{2}\right)^{15}=$$
$$\left(\frac{2e^{\frac{2}{3}\pi i}}{2e^{oi}}\right)^{15}  + \left(\frac{2e^{-\frac{2}{3}\pi i}}{2e^{oi}}\right)^{15}=$$
$$(e^{\frac{2}{3}\pi i})^{15}+(e^{-\frac{2}{3}\pi i})^{15}=$$
$$e^{10\pi i}+e^{-10\pi i}=2$$
