Cauchy-Riemann Equations/Expression of Derivative

Theorem
Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.

Let $f: D \to \C$ be a complex function on $D$.

Let $u, v: \left\{ {\left({x, y}\right) \in \R^2: x + i y = z \in D }\right\} \to \R$ be two real-valued functions defined as:


 * $u \left({x, y}\right) = \operatorname{Re} \left({f \left({z}\right) }\right)$


 * $v \left({x, y}\right) = \operatorname{Im} \left({f \left({z}\right) }\right)$

where:
 * $\operatorname{Re} \left({f \left({z}\right)}\right)$ denotes the real part of $f \left({z}\right)$
 * $\operatorname{Im} \left({f \left({z}\right)}\right) $ denotes the imaginary part of $f \left({z}\right)$.

Then $f$ is complex-differentiable in $D$ :


 * $u$ and $v$ are differentiable in their entire domain

and:
 * The following two equations, known as the Cauchy-Riemann equations, hold for the partial derivatives of $u$ and $v$:


 * $(1): \quad \dfrac {\partial u} {\partial x} = \dfrac {\partial v} {\partial y}$
 * $(2): \quad \dfrac {\partial u} {\partial y} = - \dfrac {\partial v} {\partial x}$

If the conditions are true, then for all $z \in D$:


 * $f' \left({z}\right) = \dfrac{\partial f}{\partial x} \left({z}\right) = -i \dfrac{\partial f}{\partial y} \left({z}\right)$

Proof
Let $z = x + i y$.

Then:

Similarly: