Integrate $\displaystyle\oint_{_C}\frac{1}{8z^{3}-1}dz$ where $C=|z|=1$
using $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})\text{ with } a=2z\text{, }b=1$ the denominator is rewritten to $(2z-1)(4z^{2}+2z+1)$ and the roots are found to be:
$\displaystyle z_{1}=\frac{1}{2}$
$\displaystyle z_{2}=\frac{-1}{4}+i\frac{\sqrt{3}}{4}$,
$\displaystyle z_{3}=\frac{-1}{4}-i\frac{\sqrt{3}}{4}$
When I take $|z_{i}|$, for $i=1,2,3$, I find $\displaystyle |z_{i}|=\frac{1}{2}$
So from here I find $2\pi i\sum\limits_{i=1}^{3}Res(f(z),z_{i})$, because all of the roots are inside $|z|=1$ and the result I get is that the sum of the Residues equals zero, so the integral equals zero.
However, the reason for this post, I am being told that only $z_{1}$ lies within $|z|=1$, and I do not understand how this is the case. If it is the only zero within $C$ can someone explain why the other two zeros are NOT within $C$?