[proofplan] We subtract the principal parts of the Laurent series at each singularity $z_j$ to obtain a function $g$ holomorphic on all of $U$. Cauchy's theorem on the simply connected domain $U$ kills $\int_\gamma g \, dz$. For each principal part, the terms $(z - z_j)^n$ with $n \leq -2$ have antiderivatives on $\mathbb{C} \setminus \{z_j\}$ and integrate to zero over closed paths. The surviving $c_{-1}^{(j)}/(z - z_j)$ terms are evaluated via the [Winding Number Integral Formula](/theorems/351). [/proofplan] [step:Subtract principal parts to isolate a holomorphic function on $U$] For each $j = 1, \ldots, k$, let \begin{align*} f(z) = \sum_{n=-N_j}^{\infty} c_n^{(j)} (z - z_j)^n \end{align*} be the [Laurent series expansion](/theorems/350) of $f$ near $z_j$. Define the principal part at $z_j$: \begin{align*} P_j(z) = \sum_{n=-N_j}^{-1} c_n^{(j)} (z - z_j)^n. \end{align*} Each $P_j$ is holomorphic on $\mathbb{C} \setminus \{z_j\}$. The function $g = f - \sum_{j=1}^k P_j$ is holomorphic on $U \setminus \{z_1, \ldots, z_k\}$, and at each $z_j$ the Laurent expansion of $g$ has no negative-power terms (the principal part has been subtracted), so $z_j$ is a removable singularity. Therefore $g$ extends to a holomorphic function on $U$. [/step] [step:Apply Cauchy's theorem to $g$ on the simply connected domain $U$] Since $U$ is simply connected and $g$ is holomorphic on $U$, [Cauchy's theorem for simply connected domains](/theorems/344) gives \begin{align*} \int_\gamma g(z) \, dz = 0. \end{align*} Therefore \begin{align*} \int_\gamma f(z) \, dz = \sum_{j=1}^k \int_\gamma P_j(z) \, dz. \end{align*} [/step] [step:Evaluate each principal part integral using the winding number formula] For each $j$, consider $\int_\gamma P_j(z) \, dz$. The terms $(z - z_j)^n$ with $n \leq -2$ have antiderivatives on $\mathbb{C} \setminus \{z_j\}$: namely $(z - z_j)^{n+1}/(n+1)$, which is holomorphic on $\mathbb{C} \setminus \{z_j\}$. Since $\gamma$ is a closed path in $\mathbb{C} \setminus \{z_j\}$, these integrals vanish by the Fundamental Theorem of Contour Integration. The only surviving term is $n = -1$: \begin{align*} \int_\gamma P_j(z) \, dz = c_{-1}^{(j)} \int_\gamma \frac{dz}{z - z_j}. \end{align*} By the [Winding Number Integral Formula](/theorems/351), $\int_\gamma \frac{dz}{z - z_j} = 2\pi i \, I(\gamma, z_j)$. Since $c_{-1}^{(j)} = \operatorname{Res}(f, z_j)$: \begin{align*} \int_\gamma P_j(z) \, dz = 2\pi i \, I(\gamma, z_j) \, \operatorname{Res}(f, z_j). \end{align*} [guided] Why do the terms $(z - z_j)^n$ with $n \leq -2$ integrate to zero? The function $(z - z_j)^{n+1}/(n+1)$ is well-defined and holomorphic on $\mathbb{C} \setminus \{z_j\}$ for $n \leq -2$ (since $n + 1 \leq -1 \neq 0$). The Fundamental Theorem of Contour Integration states that if $F$ is an antiderivative of $h$ on a domain, then $\int_\gamma h \, dz = F(\gamma(b)) - F(\gamma(a)) = 0$ for any closed path $\gamma$. The term $n = -1$ is different: $1/(z - z_j)$ does not have an antiderivative on $\mathbb{C} \setminus \{z_j\}$ (its "antiderivative" $\log(z - z_j)$ is multi-valued). This is precisely what the winding number captures: the integral $\int_\gamma \frac{dz}{z - z_j} = 2\pi i \, I(\gamma, z_j)$ measures the multi-valuedness of $\log(z - z_j)$ along $\gamma$. [/guided] [/step] [step:Sum over all singularities to obtain the residue formula] Summing over $j = 1, \ldots, k$: \begin{align*} \int_\gamma f(z) \, dz = \sum_{j=1}^k 2\pi i \, I(\gamma, z_j) \, \operatorname{Res}(f, z_j), \end{align*} and dividing by $2\pi i$: \begin{align*} \frac{1}{2\pi i} \int_\gamma f(z) \, dz = \sum_{j=1}^k I(\gamma, z_j) \, \operatorname{Res}(f, z_j). \end{align*} [/step]