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

(2)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)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)