polar coordinates and complex numbers Prove that $\dfrac{-1+i\sqrt{3}}{2}$ is a cube root of $1$. 
I believe I must use polar coordinates to solve this.  Perhaps  $z=r\cos(\theta)+i\sin (\theta)$.  Any help would be great!  

Hint: 
$x$ is a cubic root of $y$ if and only if $x^3 = y$. So see what $x^3$ is. No need to use what you call polar coordinates. 

EDIT: $$ \left({\dfrac{-1+i\sqrt{3}}{2}}\right)^3 = \dfrac{(-1)^3 + 3(-1)^2( i\sqrt{3} ) + 3(-1)( i\sqrt{3})^2 + (i\sqrt{3})^3}{8} =  \dfrac{-1 + 3 \sqrt 3 \cdot i + 9 - 3 \sqrt 3 \cdot i }{8} = 1 $$
