Show that \[ \int_{S^1}z^k \dz = \begin{cases} 2\pi i & k=-1 \\ 0 & \text{else}. \end{cases} ,\] and thus \[ \int \sum_{k\geq -M} c_k z^k = \sum_{k\geq -M} \int c_k z^k = 2\pi i c_{-1} ,\] i.e. the integral picks out the \(c_{-1}\) coefficient in a Laurent series expansion.
Solution
\[ \int_\gamma z^k \dz = \int_0^{2\pi} e^{ik\theta} ie^{i\theta } \dtheta = i\int_0^{2\pi} e^{i(k+1)\theta \dtheta } = \begin{cases} 2\pi i & k=-1 \\ 0 & \text{else}. \end{cases} \]