Cauchy-Riemann Equations

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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 $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ 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 $D$ if and only if:

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


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

Necessary Condition

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:

\(\ds \map u {x + h, y + k}\) \(=\) \(\ds \map \Re {\map f z} + \map \Re {t \map {f'} z} + \map \Re {t \map \epsilon t}\) Addition of Real and Imaginary Parts
\(\ds \) \(=\) \(\ds \map u {x, y} + \map \Re {h a - k b + i h b + i a k} + \map \Re {h \map \epsilon t + i k \map \epsilon t}\) Definition of Complex Multiplication
\(\ds \) \(=\) \(\ds \map u {x, y} + h a - k b + h \map \Re {\map \epsilon t} + k \map \Re {i \map \epsilon t}\) Multiplication of Real and Imaginary Parts
\(\ds \) \(=\) \(\ds \map u {x, y} + h \paren {a + \map \Re {\map \epsilon t} } + k \paren {-b + \map \Re {i \map \epsilon t} }\)

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:

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

$\Box$


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

\(\ds \map v {x + h, y + k}\) \(=\) \(\ds \map v {x, y} + \map \Im {h a - k b + i h b + i a k} + \map \Im {h \map \epsilon t + i k \map \epsilon t}\)
\(\ds \) \(=\) \(\ds \map v {x, y} + h \paren {b + \map \Im {\map \epsilon t} } + k \paren {a + \map \Im {i \map \epsilon t} }\) by a similar calculation to the one above

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.

$\Box$


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.

$\Box$


Sufficient Condition

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:

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

$\blacksquare$


Expression of Derivative

Let $z = x + i y$.

Then:

\(\ds \dfrac {\partial f} {\partial x} \left({z}\right)\) \(=\) \(\ds \dfrac {\partial u} {\partial x} \left({x, y}\right) + i \dfrac {\partial v} {\partial x} \left({x, y}\right)\)
\(\ds \) \(=\) \(\ds \operatorname{Re} \left({f' \left({z}\right) }\right) + i \operatorname{Im} \left({f' \left({z}\right) }\right)\) from the last part of the proof for sufficient condition
\(\ds \) \(=\) \(\ds f' \left({z}\right)\)

Similarly:

\(\ds -i \dfrac {\partial f} {\partial y} \left({z}\right)\) \(=\) \(\ds -i \left({\dfrac {\partial u} {\partial y} \left({x, y}\right) + i \dfrac {\partial v} {\partial y} \left({x, y}\right) }\right)\)
\(\ds \) \(=\) \(\ds -i \left({ -\operatorname{Im} \left({f' \left({z}\right) }\right) + i \operatorname{Re} \left({f' \left({z}\right) }\right) }\right)\) from the last part of the proof for sufficient condition
\(\ds \) \(=\) \(\ds f' \left({z}\right)\)

$\blacksquare$


Source of Name

This entry was named for Augustin Louis Cauchy and Georg Friedrich Bernhard Riemann.


Historical Note

The Cauchy-Riemann Equations were used by Bernhard Riemann as the basis for his theory of the concept of the analytic function.


Sources

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