Given
\(w\in B_R(z_0)\) we can find an
\(R'> 0\) such that
\(w\in \overline{B}_{R'}(z_0)\subseteq B_R(z_0)\text{.}\) (See
Figure 1.18.2.) Since
\(f\) is analytic on
\(B_R(z_0)\text{,}\) we have
\begin{equation*}
f(w)=\frac{1}{2\pi i}\int_{\gamma_{R'}}\frac{f(z)}{z-w}\, dz\text{,}
\end{equation*}
where \(\gamma_{R'}=w+R'e^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\) Since \(R'=\abs{z-z_0}< \abs{w-z_0}\) for all \(z\in \gamma_{R'}\text{,}\) we have \(\abs{(w-z_0)/(z-z_0)}< 1\text{,}\) and hence
\begin{align*}
\frac{1}{z-w} \amp =\frac{1}{z-z_0-(w-z_0}\\
\amp =\frac{1}{z-z_0}\cdot \frac{1}{1-(w-z_0)/(z-z_0)}\\
\amp = \frac{1}{z-z_0}\sum_{n=0}^\infty ((w-z_0)/(z-z_0))^n\\
\amp = \sum_{n=0}^\infty \frac{1}{(z-z_0)^{n+1}}(w-z_0)^n
\end{align*}
for all \(z\in \gamma_{R'}\text{.}\) We then have
\begin{align*}
f(w)\amp =\frac{1}{2\pi i}\int_{\gamma_{R'}}\frac{f(z)}{z-w}\, dz \\
\amp = \frac{1}{2\pi i}\int_{\gamma_{R'}} \left(\sum_{n=0}^\infty \frac{f(z)}{(z-z_0)^{n+1}}(w-z_0)^n\right) \, dz \\
\amp =\frac{1}{2\pi i}\sum_{n=0}^\infty \int_{\gamma_{R'}}\frac{1}{(z-z_0)^{n+1}}(w-z_0)^n\, dz \\
\amp = \frac{1}{2\pi i}\sum_{n=0}^\infty \left(\int_{\gamma_{R'}}\frac{1}{(z-z_0)^{n+1}}\, dz\right)(w-z_0)^n\\
\amp = \sum_{n=0}^\infty a_n(w-z_0)^n\text{,}
\end{align*}
where
\begin{equation*}
a_n=\frac{1}{2\pi i}\int_{\gamma_{R'}}\frac{1}{(z-z_0)^{n+1}}\, dz\text{.}
\end{equation*}
Of course, there is one subtle detail to consider here: namely, why are we justified in swapping the order of the integral and infinite sum? The answer, as usual, hinges on a uniformly convergent sequence. In this case, letting
\begin{equation*}
f_k(z)=\frac{f(z)}{(z-z_0)}g_k(z), g_k(z)=\sum_{n=0}^k\left((w-z_0)/(z-z_0)\right)^n\text{.}
\end{equation*}
We claim that
\begin{equation*}
f_k\to \frac{f(z)}{w-z_0}=\frac{f(z)}{z-z_0}\cdot \frac{1}{1-(w-z_0)/(z-z_0)}
\end{equation*}
uniformly on \(\gamma_{R'}\text{.}\) Indeed, since \(g(z)=\frac{f(z)}{(z-z_0)}\) is continuous on \(\gamma_{R'}\text{,}\) we have (using the extreme value theorem) \(\abs{g(z)}\leq M\) for some \(M\geq 0\text{.}\) Furthermore, since power series satisfy a uniform convergence property, we know that \(g_k\to \frac{1}{1-(w-z_0)/(z-z_0)}\) uniformly. Given any \(\epsilon > 0\text{,}\) we can thus pick an integer \(N\geq 0\) such that \(\abs{g_k-\tfrac{1}{1-(w-z_0)/(z-z_0)}}< \tfrac{\epsilon}{M}\) for all \(k> N\) and \(z\in \gamma_{R'}\text{.}\) It follows that
\begin{align*}
\abs{f_k(z)- \frac{f(z)}{w-z_0}} \amp = \abs{f_k(z)- \frac{f(z)}{z-z_0}\cdot \frac{1}{1-(w-z_0)/(z-z_0)}}\\
\amp =\abs{\frac{f(z)}{z-z_0}}\abs{g_k(z)-\frac{1}{1-(w-z_0)/(z-z_0)}}\\
\amp < M\cdot \epsilon/M\\
\amp = \epsilon
\end{align*}
for all \(k> N\) and \(z\in \gamma_{R'}\text{.}\) This proves our uniform convergence claim, and thus justifies our bringing the integral into the infinite sum.
Lastly, we observe that the derivative formula for the power series coefficients
\(a_n\) follows from
Corollary 1.17.15. As for the stated integral formula for
\(a_n\text{,}\) by the deformation of paths prinicple, we have for any
\(r\in (0,R)\)
\begin{equation*}
a_n=\frac{1}{2\pi i}\int_{\gamma_{R'}}\frac{f(z)}{z-z_0}\, dz=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-z_0}\, dz\text{,}
\end{equation*}
where \(\gamma(t)=z_0+re^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\) This completes the proof.
