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:

$f'(z_{0}):=\lim _{z\to z_{0}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}$

The ratio and limit are evaluated as complex numbers.

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:

$f'(z_{0})={\frac {\partial u}{\partial x}}(z_{0})+i{\frac {\partial v}{\partial x}}(z_{0})$

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:

$f'(z_{0})={\frac {\partial v}{\partial y}}(z_{0})-i{\frac {\partial u}{\partial y}}(z_{0})$

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:

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