why is $e^{\frac{\pi i}{3}}$ a root of $f(x) = x^{2}-x+1$? I am not sure how this comes about since, $f(e^{\frac{\pi i}{3}})=(e^{2 \pi i})^{\frac{1}{3}}-(e^{\pi i})^{\frac{1}{3}}+1 = (1)^{\frac{1}{3}}-(-1)^{\frac{1}{3}}+1 = 1+1+1 = 3$

$(z^a)^b \ne (z^b)^a$ if $a,b$ aren't both integers.  
Indeed $z^{\frac 1n}$ is not well (or usefull) defined at all.  There are $n$ values of $w$ so that $w^3 = z$ so if we decide arbitrarily that one is correct and the other $n-1$ aren't we can do anything because if we have $w^n = z$ we can't play God and assume that one of them is right and the others aren't.
$e^{\frac {2\pi i}n} := \cos \frac {2\pi}n + i\sin \frac {2\pi}n$.
It is NOT true that $e^{\frac{2\pi i}n} = (e^{2\pi i})^{\frac 13}$.  And although $e^{2\pi i} =1$.  And $1^{\frac 13}$ is not well defined and it doesn't make any sense to say "$1^{\frac 13} =1$" when talking of complex numbers.
$f(e^{\frac {2\pi i}3}) = (e^{\frac {2\pi i}3})^2 -e^{\frac {2\pi i}3} + 1=$
$e^{\frac {4\pi i}3}-e^{\frac {2\pi i}3} + 1=$
$(\cos \frac {4\pi}3 + i\sin \frac {4\pi}3) - (\cos \frac {2\pi}3 + i\sin \frac {2\pi}3) + 1=$
$(-\frac 12-i\frac {\sqrt 3}2)-(-\frac 12+i\frac {\sqrt 3}2) + 1 = $
$(-\frac 12 -\frac 12) + (-\frac {\sqrt 3}2 + \frac {\sqrt 3}2)i + 1=$
$-1 + 0i + 1 = 0$.
