The residue theorem

Step 1 (Deform the contour). By Cauchy's integral theorem, $\oint_\gamma f(z)\,dz = 0$ for any closed curve bounding a region where $f$ is holomorphic. Since $f$ fails to be holomorphic only at $z_1,\dots,z_n$, surround each $z_k$ with a tiny positively oriented circle $C_k$ of radius $\varepsilon$ small enough that the circles are disjoint and lie inside $\gamma$. Cutting slits from $\gamma$ to each $C_k$ turns the region between $\gamma$ and the $C_k$ into a simply connected domain where $f$ is holomorphic, so the integral over the boundary of that region is $0$; the slit contributions cancel in pairs, leaving $\oint_\gamma f(z)\,dz = \sum_{k} \oint_{C_k} f(z)\,dz$.

Step 2 (Evaluate each small circle). Fix $k$ and expand $f$ as its Laurent series $f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$ around $z_k$, valid on the punctured disk containing $C_k$. Every term $a_n(z-z_k)^n$ with $n \neq -1$ has an antiderivative single-valued on the punctured disk, so it integrates to $0$ around the closed circle $C_k$; only the term $a_{-1}(z-z_k)^{-1}$ survives.

Step 3 (Compute the surviving integral). Parametrize $C_k$ by $z = z_k + \varepsilon e^{i\theta}$ for $\theta \in [0, 2\pi]$, so $dz = i\varepsilon e^{i\theta}\,d\theta$ and $\dfrac{dz}{z-z_k} = i\,d\theta$. Then $\oint_{C_k} \dfrac{a_{-1}}{z-z_k}\,dz = a_{-1} \int_0^{2\pi} i\,d\theta = 2\pi i\, a_{-1} = 2\pi i\, \mathrm{Res}(f, z_k)$.

Step 4 (Sum up). Substituting Step 3 into the identity from Step 1 gives $\oint_\gamma f(z)\,dz = 2\pi i \sum_{k} \mathrm{Res}(f, z_k)$, which is exactly the residue theorem.