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.
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.)