Evaluate $\oint_C \dfrac{dz}{z^3 - 1}$, where $C$ is in the circle $|z + 1| = \dfrac{3}{2}$: Discrepancy Between Solutions. I have the following complex analysis contour integration problem:

Evaluate $\oint_C \dfrac{dz}{z^3 - 1}$, where $C$ is in the circle $|z + 1| = \dfrac{3}{2}$.

There is a discrepancy between my solution and the solution provided by the instructor.
The instructor's solution is as follows:

$\oint_C \dfrac{dz}{z^3 - 1} = 2\pi i \left( \dfrac{-\sqrt{3} + 3i}{6}  +  \dfrac{-\sqrt{3} - 3i}{6} \right) = - \dfrac{2 \pi i}{\sqrt{3}}$

My solution is as follows:
$\dfrac{1}{z^3 - 1} = \dfrac{1}{(z - 1)\left( z + \dfrac{1}{2} - \dfrac{i \sqrt{3}}{2} \right) \left( z + \dfrac{1}{2} + \dfrac{i \sqrt{3}}{2} \right)}$
Therefore, we have poles of order $1$ at $z = 1$, $z = -\dfrac{1}{2} + \dfrac{i \sqrt{3}}{2}$, $z = -\dfrac{1}{2} - \dfrac{i \sqrt{3}}{2}$. However, the pole at $z = 1$ is outside of $C$, and so we exclude it since its integral will equal $0$.
The residue theorem states that:

$\oint_C f(z) \ dz = 2\pi i \sum_{k = 1}^m res(f, c_k)$

The residue is calculated as follows:

Suppose that $f$ has a pole of order $m$ at $c$, so that
$res(f, c) = \dfrac{1}{(m - 1)!} g^{m - 1} (c)$,
where $g(z) = (z - c)^m f(z)$

We get $res\left[ \dfrac{1}{(z - 1)\left( z + \dfrac{1}{2} - \dfrac{i \sqrt{3}}{2} \right) \left( z + \dfrac{1}{2} + \dfrac{i \sqrt{3}}{2} \right)} , \dfrac{-1}{2} + \dfrac{i \sqrt{3}}{2} \right] = \dfrac{-2}{3\sqrt{3}i + 3}$
and
$res\left[ \dfrac{1}{(z - 1)\left( z + \dfrac{1}{2} - \dfrac{i \sqrt{3}}{2} \right) \left( z + \dfrac{1}{2} + \dfrac{i \sqrt{3}}{2} \right)} , \dfrac{-1}{2} - \dfrac{i \sqrt{3}}{2} \right] = \dfrac{2}{3\sqrt{3}i - 3}$
Therefore, by the residue theorem, we get
$\oint_C \dfrac{1}{z^3 - 1} = 2\pi i \left( \dfrac{2}{3\sqrt{3}i - 3} - \dfrac{2}{3\sqrt{3}i + 3}\right)$
$= \dfrac{-2\pi i}{3}  $
I would greatly appreciate it if people could please take the time to check my solution and clarify any errors.
 A: First of all, what you say the instructor has makes no sense and the arithmetic in it is not correct. What is missing is a sum (with the $2\pi i$ factored out). Note that your residue at the primitive cube root of unity $\omega = -\dfrac12+i\dfrac{\sqrt3}2$ is $\dfrac 1{3\omega^2} = \dfrac13 \omega$. Adding this to its conjugate makes the arithmetic a lot easier.
EDIT: So there's no difference between the solutions, except the instructor is using a more convenient computation of the residue, as I did. The instructor probably should have included a bit more detail. (That said, I hate using the definition of residue that you are using, although I realize it is widely taught. I far prefer to use Laurent series, when needed, along with the observation that
when $h$ has a simple zero at $z=a$ and $g$ is holomorphic at $z=a$, then we have
$$\text{Res}_{z=a} \frac{g(z)}{h(z)} = \frac{g(a)}{h'(a)}.$$
This should be immediate from your approach and is certainly immediate from thinking about the Laurent series.)
