Cauchy-Riemann Equations/Sufficient Condition

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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 Epsilon-Function 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 real functions that converge to $0$ as $h$ and $k$ tend to $0$.

We have by hypothesis that the partial derivatives are continuous.

Then Epsilon-Function Differentiability Condition shows that $\epsilon_{u x}, \epsilon_{u y}, \epsilon_{v x}, \epsilon_{v y}$ are continuous real functions.

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

\(\ds \map f {z + t}\) \(=\) \(\ds \map u {x + h, y + k} + i \map v {x + h, y + k}\)
\(\ds \) \(=\) \(\ds \map u {x, y} + i \map v {x, y} + h \paren {a + i b + \map {\epsilon_{u x} } h + i \map {\epsilon_{v x} } h} + k \paren {i a - b + \map {\epsilon_{u y} } k + i \map {\epsilon_{v y} } k}\)
\(\ds \) \(=\) \(\ds \map f z + t \paren {a + i b + \dfrac h t \map {\epsilon_{u x} } h + \dfrac h t i \map {\epsilon_{v x} } h - \dfrac k t i \map {\epsilon_{u y} } k + \dfrac k t \map {\epsilon_{v y} } k}\)

Now, define $\map \epsilon t$ as:

\(\ds \map \epsilon t\) \(=\) \(\ds \dfrac h t \map {\epsilon_{u x} } h + \dfrac h t i \map {\epsilon_{v x} } h - \dfrac k t i \map {\epsilon_{u y} } k + \dfrac k t \map {\epsilon_{v y} } k\)
\(\ds \leadsto \ \ \) \(\ds \cmod {\map \epsilon t}\) \(\le\) \(\ds \cmod {\dfrac h t \map {\epsilon_{u x} } h} + \cmod {\dfrac h t \map {\epsilon_{v x} } h} + \cmod {\dfrac k t \map {\epsilon_{u y} } k} + \cmod {\dfrac k t \map {\epsilon_{v y} } k}\) Triangle Inequality for Complex Numbers
\(\ds \) \(=\) \(\ds \cmod {\dfrac h t} \cmod {\map {\epsilon_{u x} } h} + \cmod {\dfrac h t} \cmod {\map {\epsilon_{v x} } h} + \cmod {\dfrac k t} \cmod {\map {\epsilon_{u y} } k} + \cmod {\dfrac k t} \cmod {\map {\epsilon_{v y} } k}\) Complex Modulus of Product of Complex Numbers


By Modulus Larger than Real Part and Imaginary Part:

$\cmod k \le \cmod t$

and:

$\cmod h \le \cmod t$

Hence dividing by $\cmod t$ we have:

$\cmod {\dfrac h t} \le 1$

and:

$\cmod {\dfrac k t} \le 1$

From Composite of Continuous Mappings is Continuous, it follows that $\map \epsilon t$ is continuous.

Thus:

\(\ds \lim_{t \mathop \to 0} \cmod {\map \epsilon t}\) \(\le\) \(\ds \lim_{t \mathop \to 0} \cmod {\dfrac h t} \cmod {\map {\epsilon_{u x} } h} + \cmod {\dfrac h t} \cmod {\map {\epsilon_{v x} } h}\) Combined Sum Rule for Limits of Complex Functions
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \cmod {\dfrac k t} \cmod {\map {\epsilon_{u y} } k} + \cmod {\dfrac k t} \cmod {\map {\epsilon_{v y} } k}\)
\(\ds \) \(\le\) \(\ds \lim_{t \mathop \to 0} \cmod {\map {\epsilon_{u x} } h} + \lim_{t \mathop \to 0} \cmod {\map {\epsilon_{v x} } h}\) as $\cmod {\dfrac h t} \le 1$ and $\cmod {\dfrac k t} \le 1$ from above
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \lim_{t \mathop \to 0} \cmod {\map {\epsilon_{u y} } k} + \lim_{t \mathop \to 0} \cmod {\map {\epsilon_{v y} } k}\)
\(\ds \) \(=\) \(\ds 0\)


This shows that:

$\ds \lim_{t \mathop \to 0} \map \epsilon t = 0$

Then the Epsilon-Function Complex Differentiability Condition shows that:

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

$\blacksquare$


Sources

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