Taylor Series of Analytic Complex Functions

# Taylor Series of Analytic Complex Functions

We are already familiar with Taylor series for infinitely differentiable real-valued functions. The notion of a Taylor series for an analytic complex function is analogous.

 Definition: Let $A \subseteq \mathbb{C}$ be open and let $f : A \to \mathbb{C}$ be analytic on $A$. Let $z_0 \in A$. Then the Taylor Series Expansion of $f$ Centered at $z_0$ is $\displaystyle{T_f^{z_0} (z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0}{n!} (z - z_0)^n}$.

If $f$ is analytic on $A$ then we know that the derivatives of $f$ of all orders exist, so $T_f^{z_0}$ is a valid formal sum. Whether or not $T_f^{z_0}$ converges will be discussed shortly.

For example, consider the function $f(z) = e^z$ which is analytic on all of $\mathbb{C}$. Let's compute the Taylor series expansion of $f$ centered at $z_0 = 0$. We know that $f^{(n)}(z) = e^z$ for all $n \in \{ 0, 1, 2, ... \}$. Therefore, for each $n \in \{0, 1, 2, ... \}$ we have that:

(1)
\begin{align} \quad \frac{f^{(n)}(0)}{n!} = \frac{e^0}{n!} = \frac{1}{n!} \end{align}

Thus the Taylor series expansion of $f(z) = e^z$ centered at $z_0 = 0$ is:

(2)
\begin{align} \quad T_f^{0} (z) = \sum_{n=0}^{\infty} \frac{z^n}{n!} \end{align}

For another example, consider the function $f(z) = \frac{1}{1 - z}$. This function is analytic on $\mathbb{C} \setminus \{ 0 \}$. Let's compute the Taylor series expansion of $f$ centered at $z_0 = 0$. We have that:

(3)
\begin{align} \quad \frac{f(0)}{0!} = f(0) = 1 \end{align}
(4)
\begin{align} \quad \frac{f^{(1)}(0)}{1!} = f'(0) = \frac{1}{(1 - 0)^2} = 1 \end{align}
(5)
\begin{align} \quad \frac{f^{(2)}(0)}{2!} = \frac{f''(0)}{2!} = \frac{2}{2!(1 - 0)^3} = 1 \end{align}

In fact, it's not hard to show that for all $n \in \{ 0 , 1, 2, ... \}$ that $\displaystyle{\frac{f^{(n)}(0)}{n!} = 1}$. So the Taylor series of $\displaystyle{f(z) = \frac{1}{1 - z}}$ centered at $z_0 = 0$ is:

(6)
\begin{align} \quad T_f^{0} (z) = \sum_{n=0}^{\infty} z^n \end{align}