Consider the function \(g\text{,}\) defined as
\begin{equation}
g(z) = \begin{cases} (z-\alpha)^2f(z) , \amp \text{ when } z \ne \alpha ; \\ \qquad 0, \amp \text{ when } z=\alpha. \end{cases}\tag{7.5.1}
\end{equation}
Clearly, \(g\) is analytic in at least \(D_r^*(\alpha)\text{.}\) Straightforward calculation yields
\begin{equation*}
g\,'(\alpha) = \lim_{z \to\alpha}\frac{g(z) -g(\alpha)}{z-\alpha}=\lim_{z \to\alpha }(z-\alpha)f(z) = 0\text{.}
\end{equation*}
The last equation follows because
\(f\) is bounded. Thus
\(g\) is also analytic at
\(\alpha\text{,}\) with
\(g(\alpha) = g\,'(\alpha) = 0\text{.}\)
By Taylor’s theorem, \(g\) has the representation
\begin{equation}
g(z) = \sum_{n=2}^{\infty}\frac{g^{(n)}(\alpha)}{n!}(z-\alpha)^n, \text{ for all } z \in D_r(\alpha)\text{.}\tag{7.5.2}
\end{equation}
We divide both sides of Equation
(7.5.2) by
\((z-\alpha)^2\) and use Equation
(7.5.1) to obtain the following power series representation for
\(f\text{:}\)
\begin{equation*}
f(z) = \sum_{n=2}^{\infty}\frac{g^{(n)}(\alpha)}{n!}(z-\alpha)^{n-2} = \sum \limits_{n=0}^{\infty}\frac{g^{(n+2)}(\alpha)}{(n+2)!}(z-\alpha)^n\text{.}
\end{equation*}
By
Theorem 4.4.7,
\(f\) is analytic at
\(\alpha\) if we define
\(f(\alpha)=\frac{g^{(2)}(\alpha)}{2!}\text{.}\) This observation completes the proof.
