Complex differential of a complex-valued function

Definition

Definition as a general limit

Suppose $U$ is an open subset of $\mathbb{C}$ and $f:U\to \mathbb{C}$ is a function. Let $z_0\in \mathbb{C}$. Then, the complex differential of $f$ at $z_0$ is given by:

$f^{\prime }(z_0):=\underset{z\to z_0}{lim}\frac{f(z)-f(z_0)}{z-z_0}$

The ratio and limit are evaluated as complex numbers.

If $f$ has a complex differential at $z_0$, we say that $f$ is complex-differentiable at $z_0$.

Definition as limits from the real and imaginary directions

If $f$ is complex-differentiable at $z_0\in \mathbb{C}$, then we can compute its derivative by using a linear direction of approach. For instance, we can look at $z=z_0+h$, where $h\in \mathbb{R}$, and take the limit as $h\to 0$. Thus, if we write:

$f(z)=u(z)+iv(z)$

where $u,v$ are real-valued functions, then we get:

$f^{\prime }(z_0)=\frac{\partial u}{\partial x}(z_0)+i\frac{\partial v}{\partial x}(z_0)$

Similarly, we can consider approach along the imaginary direction, namely, $z=z_0+ih$, where $h\in \mathbb{R}$, and let $h\to 0$. We then get:

$f^{\prime }(z_0)=\frac{\partial v}{\partial y}(z_0)-i\frac{\partial u}{\partial y}(z_0)$

It turns out that if $f$ is continuously differentiable in the real sense at $z_0$, and the two notions of differential above coincide at $z_0$, then $f$ is complex-differentiable at $z_0$, and the complex differential equals either of the expressions. The equality of the two expressions is termed the Cauchy-Riemann differential equations:

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$