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, $f(e^{\frac{\pi i}{3}})=(e^{2 \pi i})^{\frac{1}{3}}-(e^{\pi i})^{\frac{1}{3}}+1 = (1)^{\frac{1}{3}}-(-1)^{\frac{1}{3}}+1 = 1+1+1 = 3$

$x=e^{i\pi/3}=1/2+\sqrt{3}i/2.$
Then $$f(e^{i\pi/3})=[1/2+i\sqrt{3}/2]^2-[1/2+i\sqrt{3}/2]+1=1/4-3/4+i\sqrt{3}/2-1/2-i\sqrt{3}/2+1=0$$
