Complex numbers: proof of equality Can anyone help me prove following equality?
$$\left(\frac{-1+i\sqrt{3}}{2}\right)^n + \left(\frac{-1-i\sqrt{3}}{2}\right)^n$$ $$=2 \iff \frac{n}3\in \mathbb{N}$$
$$=-1 \iff \frac{n}3\not\in \mathbb{N}$$
This is what I've got:
$$\left(\frac{-1+i\sqrt{3}}{2}\right)^n + \left(\frac{-1-i\sqrt{3}}{2}\right)^n$$
$$= \left(\frac{2(cos(-\pi/3)+i sin(-\pi/3))}2\right)^n + \left(\frac{2(cos(\pi/3)+i sin(\pi/3))}2\right)^n$$
$$=[cos(-\pi/3)+isin(-\pi/3)]^n-[cos(\pi/3)+isin(\pi/3)]^n$$
$$=cos(n\pi/3)-isin(n\pi/3)-cos(n\pi/3)-isin(n\pi/3)$$
$$=-2isin(n\pi/3)$$
and if $n/3 \in \mathbb{N}$, I get $n'\pi$ and $sin(n'\pi)=0$ which isn't the result I needed...
 A: The idea is good, but your angles are not quite right: for example, $\cos{(-\pi/3)} = \cos{(\pi/3)} = 1/2$ and you want an angle $\alpha$ with $\cos{\alpha} = -1/2$ and $\sin{\alpha} = \sqrt{3}/2$. One such angle is $\alpha = 2\pi/3$, 
leading to $2\cos{(2n\pi/3)}$, which is equal to 2 if and only if $n/3$ is an integer and to $-1$ otherwise. [Note that the question also makes sense for negative integers $n$.]
A: Lets write 
$$\zeta=\frac{-1+i\sqrt{3}}{2}$$
then 
$$\zeta^{-1}=\zeta^2=\frac{-1-i\sqrt{3}}{2}$$
Now  $\zeta$ is a cube root of $1$, so $\zeta^3=1$ and also $\zeta^{-3}=1$
Thus if $3$ divides $n$ then 
$$\zeta^n+\zeta^{-n}=1+1=2$$
If $3$ does not divide $n$ then $n\equiv 1 \ \text{or} \ -1 \mod 3$ and so 
$$\zeta^n+\zeta^{-n}=\zeta+\zeta^{-1}=-1$$
A: Since
$$
\left(\frac{-1+i\sqrt{3}}{2}\right)^3=1=\left(\frac{-1-i\sqrt{3}}{2}\right)^3
$$
you know that both numbers are roots of the polynomial $x^2+x+1$.
For $n=0$ we get $2$; for $n=1$ we get $-1$. So the given expression is the solution of the recurrence
$$
a_{n+2}=-a_{n+1}-a_n,\quad a_0=2, \quad a_1=-1
$$
We have
$$
a_{n+3}=-a_{n+2}-a_{n+1}=a_{n+1}+a_n-a_{n+1}=a_n
$$
A: Note that using Euler's Formula, we can write $\frac{-1\pm i\sqrt3}{2}=\cos(2\pi/3)\pm i\sin(2\pi/3)=e^{\pm i2\pi/3}$.  
Therefore, from De Moivre's Formula, we have
$$\begin{align}
\left(\frac{-1+i\sqrt3}{2}\right)^n+\left(\frac{-1-i\sqrt3}{2}\right)^n&=e^{i2n\pi/3}+e^{-i2n\pi/3}\\\\
&=2\cos(2n\pi/3)
\end{align}$$
Finally, we see that
$$\left(\frac{-1+i\sqrt3}{2}\right)^n+\left(\frac{-1-i\sqrt3}{2}\right)^n=\begin{cases}2&,n=0,3,6\dots\\\\-1&,n\ne0,3,6\dots\end{cases}$$
A: Theres no point in converting to trig unless you are familiar and comfortable with polar coordinates.  If so:
$(\frac {-1 + i\sqrt{3}}2)^n+(\frac {-1 - i\sqrt{3}}2)^n=$
$(\cos \frac {2\pi}{3} + i \sin \frac {2\pi}{3})^n + (\cos \frac {4\pi}{3} + i \sin \frac {4\pi}{3})^n=$
$(e^{i\frac{2\pi}{3}})^n + (e^{i\frac{4\pi}{3}})^n=e^{i\frac{2n\pi}{3}} + e^{i\frac{4n\pi}{3}}=$
$(\cos \frac {2n\pi}{3} + i \sin \frac {2n\pi}{3}) + (\cos \frac {4n\pi}{3} + i \sin \frac {4n\pi}{3})=$
$3|n$ i.e. if $n = 3k$
$= \cos 2k\pi + i \sin 2k \pi + \cos 4k\pi + i\sin 4k\pi = 1 + 0i + 1 + 0i = 2$
if $3\not \mid n$ ie. if $n = 3k \pm 1$
$ = \cos (\pm \frac{2\pi}3) + i \sin (\pm \frac{2\pi}3) + \cos (\pm \frac{4\pi}3) + i \sin (\pm \frac{4\pi}3)$
$= -\frac 12 \pm i\frac{\sqrt{3}}{2} -\frac 12 \mp i \frac{\sqrt{3}}{2}= -1$
But really itd be easier to realize: $(\frac {-1 \pm i\sqrt{3}}2)^3 = 1$ and $(\frac {-1 \pm i\sqrt{3}}2)^2 = \frac {-1 \mp i\sqrt{3}}2$ [just multiply them out]so
$(\frac {-1 + i\sqrt{3}}2)^{3k + i} + (\frac {-1 - i\sqrt{3}}2)^{3k + i} = $
$(\frac {-1 + i\sqrt{3}}2)^i + (\frac {-1 - i\sqrt{3}}2)^i = 1 + 1 = 2$ if $i = 0$
$=(\frac {-1 \pm i\sqrt{3}}2) + (\frac {-1 \mp i\sqrt{3}}2)$
$= -\frac 12 - \frac 12 =-1$ if $i = 1$ or $i = 2$.
