Complex Differentiable Functions

We will now touch upon one of the core concepts in complex analysis - differentiability of complex functions baring in mind that the concept of differentiability of a complex function is analogous to that of a real function.

Definition: Let $A \subseteq C$ be open, $z_0 \in A$, and let $f : A \to \mathbb{C}$. Then $f$ is said to be Complex Differentiable at $z_0$ if the limit $\displaystyle{f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}}$ exists. The limit, $f'(z_0)$ is called the Derivative of $f$ at $z_0$ provided that this limit exists.

Note that if $h = z - z_0$ then $h \to 0$ if and only if $z \to z_0$, in which case we have that:

(1)

\begin{align} \quad f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \end{align}

Also note that in both limit definitions above - the limit exists if and only if the limit is finite and the same for all paths in $\mathbb{C}$ as $h \to 0$ (or as $z \to z_0$).

For a simple example, consider the following function:

(2)

\begin{align} \quad f(z) = z \end{align}

We will show that this function is differentiable at every $z_0 \in \mathbb{C}$. We have that:

(3)

\begin{align} \quad f'(z_0) &= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \\ &=\lim_{z \to z_0} \frac{z - z_0}{z - z_0} \\ &= \lim_{z \to z_0} 1 \\ &= 1 \end{align}

Therefore $f$ is differentiable at every point $z_0 \in \mathbb{C}$ and the derivative of $f$ at every $z_0 \in \mathbb{C}$ is $f'(z_0) = 1$.

For another example, consider the following function:

(4)

\begin{align} \quad f(z) = z^2 \end{align}

We will show that this function is differentiable at $0$. We have that:

(5)

\begin{align} \quad f'(0) &= \lim_{z \to 0} \frac{f(z) - 0}{z - 0} \\ &= \lim_{z \to 0} \frac{z^2 - 0}{z - 0} \\ &= \lim_{z \to 0} \frac{z^2}{z} \\ &= \lim_{z \to 0} z \\ &= 0 \end{align}

The following theorem tells us that if a function is complex differentiable at a point then it is continuous at that point.

Theorem 1: Let $A \subseteq \mathbb{C}$ be open, $z_0 \in A$, and $f : A \to \mathbb{C}$. If $f$ is complex differentiable at $z_0$ then $f$ is continuous at $z_0$.

• Proof: Since $f$ is complex differentiable at $z_0$, $f'(z_0)$ exists and so:

(6)

\begin{align} \quad \lim_{z \to z_0} [f(z) - f(z_0)] &= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \cdot (z - z_0) \\ &= f'(z_0) \cdot \lim_{z \to z_0} (z - z_0) \\ &= f'(z_0) \cdot 0 \\ &= 0 \end{align}

• Therefore $\displaystyle{\lim_{z \to z_0} f(z) = f(z_0)}$ and so $f$ is continuous at $z_0$. $\blacksquare$