Taylor's Theorem for Analytic Complex Functions

# Taylor's Theorem for Analytic Complex Functions

 Theorem 1 (Taylor's Theorem for Analytic Complex Functions): Let $A \subseteq \mathbb{C}$ be an open connected set and let $z_0 \in A$. Let $f : A \to \mathbb{C}$ be analytic on $A$. Then there exists an $r > 0$ such that $\displaystyle{T_f^{z_0} (z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z - z_0)^n = f(z)}$ for all $z \in D(z_0, r)$. In other words, the Taylor series of $f$ at $z_0$ converges uniformly and absolutely to $f$ for all $z \in D(z_0, r)$.
• Proof: Let $A \subseteq \mathbb{C}$ be an open and connected set and let $z_0 \in A$. Since $A$ is open and $z_0 \in A$ there exists an open disk centered at $z_0$ fully contained in $A$. Take $R > 0$ such that the closed disk $\overline{D(z_0, R)}$ is fully contained in $A$ ($\overline{D(z_0, R)} \subset A$).
• Then $\overline{D(z_0, R)}$ is a compact set and so $f$ which is analytic on $\overline{D(z_0, R)}$ (and hence continuous on $\overline{D(z_0, R)}$ attains a maximum on $\overline{D(z_0, R)}$. So $f$ is bounded on $\overline{D(z_0, R)}$, i.e., there exists an $M > 0$ such that for all $z \in \overline{D(z_0, R)}$:
(1)
• Let $r > 0$ be such that $r < R$ and consider the open disk $D(z_0, r)$. For each $n \in \{ 0, 1, 2, ... \}$ let:
(2)
\begin{align} \quad a_n = \frac{f^{(n)}(z_0)}{n!} \end{align}
• From $(*)$, since $f$ is analytic on $D(z_0, r)$ and its boundary, the circle $\mid z - z_0 \mid = r$ and $\mid f(z) \mid \leq M$ on this circle, we have by Cauchy's Derivative Inequalities that for each $n \in \{ 0, 1, 2, ... \}$ that:
(3)
\begin{align} \quad \mid f^{(k)}(z_0) \mid \leq \frac{n! M}{R^n} \end{align}
• Hence:
(4)
\begin{align} \quad \mid a_n \mid \leq \frac{M}{R^n} \end{align}
• For each $n \in \{ 0, 1, 2, ... \}$ let:
(5)
\begin{align} \quad M_n = \left ( \frac{r}{R} \right )^n \end{align}
• Then note that:
(6)
\begin{align} \quad \biggr \lvert \frac{f^{(n)}}{n!} (z - z_0)^n \biggr \rvert = \mid a_n (z - z_0)^n \mid = \mid a_n \mid \mid z - z_0 \mid^n \leq \frac{M}{R^n} \cdot r^n = M M_n \quad (**) \end{align}
• However, since $r < R$, the series $\displaystyle{\sum_{n=0}^{\infty} M_n}$ converges as a geometric series and so from $(**)$ we have by the Weierstrass M-test that $T_f(z_0)$ converges uniformly (and absolutely) on $D(z_0, r)$.
• We now show that $T_f^{z_0} (z)$ converges to $f(z)$ for all $z \in D(z_0, r)$. Let $C_r$ be the positively oriented circle centered at $z_0$ with radius $r > 0$. Then by Cauchy's Integral Formula we have that for any $z \in D(z_0, r)$ (the interior of $C_r$) that:
(7)
\begin{align} \quad f(z) &= \frac{1}{2\pi i} \int_{C_r} \frac{f(w)}{w - z} \: dw \\ &= \frac{1}{2 \pi i} \int_{C_r} \frac{f(w)}{(w - z_0) - (z - z_0)} \: dw \\ &= \frac{1}{2 \pi i} \int_{C_r} f(w) \frac{1}{w - z_0} \frac{1}{1 - \frac{z - z_0}{w - z_0}} \: dw \end{align}
• Note that $\displaystyle{ \biggr \lvert \frac{z - z_0}{w - z_0} \biggr \rvert \leq \frac{r}{R} < 1}$ and so:
(8)
\begin{align} \quad \frac{1}{w - z} \frac{1}{1 - \frac{z - z_0}{w - z_0}} = \frac{1}{w - z_0}\sum_{n=0}^{\infty} \left ( \frac{z - z_0}{w - z_0} \right )^n \end{align}
• Thus, by the uniform convergence of $T_f^{z_0} (z)$ on $C_r$ we have that:
(9)
\begin{align} \quad f(z) &= \frac{1}{2\pi i} \int_{C_r} \frac{f(w)}{w - z_0} \sum_{n=0}^{\infty} \left ( \frac{z - z_0}{w - z_0} \right )^n \: dw \\ &= \sum_{n=0}^{\infty} (z - z_0)^n \frac{1}{2 \pi i} \int_{C_r} \frac{f(w)}{(w - z_0)^{n+1}} \: dw \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z - z_0)^n \quad \mathrm{by \: Cauchy's \: Integral \: Theorem \: for \: Derivatives} \\ &= T_f^{z_0}(z) \quad \blacksquare \end{align}