easiest way to Simplify $\frac{w^{13}+w^{12}+w^{11}}{10}$ if $w= \frac{-1+i\sqrt{3}}{2}$. 
w= $\frac 1 2  (-1+i\sqrt{3})$
$( w^{13}+w^{12}+w^{11}$) 1/10
this the one of the practice question !!! is there any  easy /quick/IQ kind of way to 
simplify this ..
What I have done is : 
taking $w^{11}$ common 
$w^{11}( w^{2}+w^{}+1)$ 1/10
now what next ?? should substitute value of w ... but isn't it the long way to solve ???
 A: Notice that $\omega$ is complex cube root of unity
$$x^3-1=0$$
$$(x-1)(x^2+x+1)=0$$
since $\omega$ is complex cube root of unity, $\omega\neq1$
So, $\omega$  is root of $x^2+x+1=0\implies \omega^{2}+\omega+1=0$
$$\frac{\omega^{13}+\omega^{12}+\omega^{11}}{10}=\frac{\omega^{11}(\omega^{2}+\omega+1)}{10}=0$$
A: Note that $\cos(2\pi/3)=-1/2$ and $\sin(2\pi/3)=\sqrt{3}/2$.  
Hence 
$$\begin{align}w&=\frac{-1+ i\sqrt3}2=\cos\frac{2\pi}3+i\sin\frac{2\pi}3=e^{i2\pi/3}\\
w^3&=e^{i2\pi}=1\\
w^{12}&=(w^3)^4=1\\
\frac{w^{13}+w^{12}+w^{11}}{10}&=\frac{w^{12}(w+1+w^{-1})}{10}\\
&=\frac{1(\color{red}{e^{i2\pi/3}}+1+\color{red}{e^{-i2\pi/3}})}{10}\\
&=\frac{1(\color{red}{-1}+1)}{10}\\
&=0\\\qquad \blacksquare \end{align}$$
A: From $w = (-1 + i\sqrt{3})/2$, we have
$$\begin{align}
(2w + 1)^{2} &= (i\sqrt{3})^{2}\\
w^{2} + w + 1 &= 0.
\end{align}$$
A: In general it might be an idea to break high powers into powers of powers, e.g.:
$$w^{11}=(w^3)^3\cdot w^2.$$
Another good idea, especially in this case, is to use the exponential (or trig) form for $w$. In this case:
$$w=e^{2i\frac{\pi}{3}},$$
which immediately implies:
$$w^3=e^{i\cdot2\pi}=1,$$
which means you can read exponents of $w$ modulo 3 and:
$$w^{11}=w^2=e^{i\cdot\frac{4}{3}\pi}=e^{-i\cdot\frac{2}{3}\pi}=-\frac12-i\frac{\sqrt3}{2},$$
which if looked at well enough will give you $w^2+w+1=0$ btw.
I know many other answers use more ore less the same results, but I wanted to give some general advice for similar problems.
