Cauchy-Riemann Equations/Sufficient Condition

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


 * $\map u {x, y} = \map \Re {\map f z}$


 * $\map v {x, y} = \map \Im {\map f z}$

where:
 * $\map \Re {\map f z}$ denotes the real part of $\map f z$
 * $\map \Im {\map f z}$ denotes the imaginary part of $\map f z$.

Let:


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

and:


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


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

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

and:
 * for all $z \in D$:


 * $\map {f'} z = \map {\dfrac {\partial f} {\partial x} } z = -i \map {\dfrac {\partial f} {\partial y} } z$

Proof
Suppose that the Cauchy-Riemann equations hold for $u$ and $v$ in their entire domain.

Let $h, k \in \R \setminus \set 0$, and put $t = h + i k \in \C$.

Let $\tuple {x, y} \in \R^2$ be a point in the domain of $u$ and $v$.

Put:
 * $a = \map {\dfrac {\partial u} {\partial x} } {x, y} = \map {\dfrac {\partial v} {\partial y} } {x, y}$

and:
 * $b = -\map {\dfrac {\partial u} {\partial y} } {x, y} = \map {\dfrac {\partial v} {\partial x} } {x, y}$

From the Alternative Differentiability Condition, it follows that:


 * $\map u {x + h, y} = \map u {x, y} + h \paren {a + \map {\epsilon_{u x} } h}$
 * $\map u {x, y + k} = \map u {x, y} + k \paren {-b + \map {\epsilon_{u y} } k}$
 * $\map v {x + h, y} = \map v {x, y} + h \paren {b + \map {\epsilon_{v x} } k}$
 * $\map v {x, y + k} = \map v {x, y} + k \paren {a + \map {\epsilon_{v y} } h}$

where $\epsilon_{u x}, \epsilon_{u y}, \epsilon_{v x}, \epsilon_{v y}$ are continuous real functions that converge to $0$ as $h$ and $k$ tend to $0$.

With $z = x + i y$, it follows that:

With the assumption that the partial derivatives are continuous, we have $a + i b$,, which is a function of h, converges to itself by taking h -> 0.

Thus:

Thus:

This shows that:
 * $\ds \lim_{t \mathop \to 0} \map \epsilon t = 0$

From Continuity of Composite Mapping, it follows that $\map \epsilon t$ is continuous.

Then the Alternative Differentiability Condition shows that:


 * $\map {f'} x = a + i b$