For each \(1\leq k\leq\) we have a Laurent expansion
\begin{align}
f(z) \amp = \sum_{n=1}^\infty a_{n}^k(z-z_0)^{-n}+\sum_{n=0}^\infty b_n^k(z-z_0)^n, \ 0< \abs{z-z_0} < R_k\text{.}\tag{1.92}
\end{align}
We double-index the coefficients here as they depend also on \(z_k\text{.}\) For each \(k\text{,}\) let
\begin{equation*}
g_k(z)=\sum_{n=1}^\infty a_{n}^k(z-z_k)^{-n}\text{,}
\end{equation*}
the so-called
principal part of the Laurent expansion of
\(f\) at
\(z_k\text{.}\) It follows from the definition of convergence of Laurent series, that since the series expansion
(1.92) converges for all
\(0< \abs{z-z_k} < R_k\text{,}\) the principal part
\(g_k(z)\) converges for all
\(z\ne z_k\text{.}\) It follows
\(g_k(z)\) extends to an analytic function on
\(U-\{z_k\}\text{,}\) and further that the function
\begin{equation*}
h(z)=f(z)-\sum_{k=1}^n h_k(z)
\end{equation*}
is analytic on \(U\) except at each \(z_k\text{,}\) where it has removable singularities. The function \(h\) can thus be extended to an analytic function on \(U\text{.}\) Since \(U\) is elementary and \(\gamma\) is closed, we have
\begin{align*}
0 \amp = \int_\gamma h\, dz\\
\amp =\int_\gamma f\, dz-\sum_{k=1}^n \int_\gamma h_k\, dz \text{.}
\end{align*}
Lastly, since \(\int_\gamma \frac{c}{(z-z_k)^n}=0\) for all \(c\in \C\text{,}\) for all integers \(n\geq 2\text{,}\) and for all \(1\leq k\leq n\) (use an antiderivative!), we have
\begin{align*}
\int_\gamma f\, dz \amp = \sum_{k=1}^n \int_\gamma h_k\, dz \\
\amp = \sum_{k=1}^n \int_\gamma \frac{a_{1}^k}{(z-z_k)}\, dz \\
\amp = 2\pi i\sum_{k=1}^n a_{1}^k\chi_\gamma(z_k)\\
\amp = 2\pi i\sum_{k=1}^n\chi_\gamma(z_k)\res_f(z_k)\text{,}
\end{align*}
as desired.
