Cauchy-Riemann Equations/Necessary 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 $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ 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 $\map f z$.

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 $z \in D$:


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

Proof
Let $z = x + i y \in D$.

The Epsilon-Function Complex Differentiability Condition shows that there exists $r \in \R_{>0}$ such that for all $t \in \map {B_r} 0 \setminus \set 0$:


 * $(\text i) \quad \map f {z + t} = \map f z + t \paren {\map {f'} z + \map \epsilon t}$

where:
 * $\map {B_r} 0$ denotes an open ball of $0$
 * $\epsilon: \map {B_r} 0 \setminus \set 0 \to \CC$ is a function with $\ds \lim_{t \mathop \to 0} \map \epsilon t = 0$.

Define $a, b, h, k \in \R$ by $a + i b = \map {f'} z$, and $h + i k = t$.

By taking the real parts of both sides of equation $(\text i)$, it follows that:

To find the partial derivative $\dfrac {\partial u} {\partial x}$, assume that $y$ is fixed, and let $t$ be wholly real.

Then $t = h$, and $k = 0$, so:


 * $\map u {x + h, y} = \map u {x, y} + h \paren {a + \map \Re {\map \epsilon h} }$

From Limits of Real and Imaginary Parts, it follows that:
 * $\ds \lim_{h \mathop \to 0} \map \Re {\map \epsilon h} = \map \Re {\lim_{h \mathop \to 0} \map \epsilon h} = \map \Re 0 = 0$

Then the Epsilon-Function Differentiability Condition proves that:
 * $\map {\dfrac {\partial u} {\partial x} } {x, y} = a$

To find the partial derivative $\dfrac {\partial u} {\partial y}$, assume that $x$ is fixed, and let $t$ be wholly imaginary.

Then $t = i k$, and $h = 0$, so:


 * $\map u {x, y + k} = \map u {x, y} + k \paren {-b + \map \Re {i \map \epsilon {i k} } }$

From Limits of Real and Imaginary Parts, it follows that:
 * $\ds \lim_{k \mathop \to 0} \map \Re {i \map \epsilon k} = \map \Re {i \lim_{k \mathop \to 0} \map \epsilon k} = 0$

Then the Epsilon-Function Differentiability Condition proves that:
 * $\map {\dfrac {\partial u} {\partial y} } {x, y} = -b$

By taking the imaginary parts of both sides of equation $(\text i)$, it follows that:

To find the partial derivative $\dfrac {\partial v} {\partial x}$, assume that $y$ is fixed, and let $t$ be wholly real.

Then $t = h$, and $k = 0$, so:


 * $\map v {x + h, y} = \map v {x, y} + h \paren {b + \map \Im {\map \epsilon h} }$

A similar argument to the ones above shows that the Epsilon-Function Differentiability Condition can be applied to prove that:


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

This proves the second Cauchy-Riemann equation.

To find the partial derivative $\dfrac {\partial v} {\partial y}$, assume that $x$ is fixed, and let $t$ be wholly imaginary.

Then $t = ik$, and $h = 0$, so:


 * $\map v {x, y + k} = \map v {x, y} + k \paren {a + \map \Im {i \map \epsilon {i k} } }$

Here, the Epsilon-Function Differentiability Condition shows that:


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

This proves the first Cauchy-Riemann equation.

From Holomorphic Function is Continuously Differentiable, it follows that $f'$ is continuous.

From Composite of Continuous Mappings is Continuous and Real and Imaginary Part Projections are Continuous, it follows that $\map \Re f$ and $\map \Im f$ are continuous.

Then all four partial derivatives are continuous, as:


 * $\map {\dfrac {\partial u} {\partial x} } {x, y} = \map {\dfrac {\partial v} {\partial y} } {x, y} = a = \map \Re {\map f z}$


 * $\map {\dfrac {\partial u} {\partial y} } {x, y} = -\map {\dfrac {\partial v} {\partial x} } {x, y} = -b = -\map \Im {\map f z}$

By definition of differentiablity of real-valued functions, it follows that $u$ and $v$ are differentiable.