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 $u, v: \left\{ {\left({x, y}\right) \in \R^2: x + i y = z \in D }\right\} \to \R$ 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 $D$ 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 $z \in D$:

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


Proof

Necessary Condition

Let $z = x+iy \in D$.

The Alternative Differentiability Condition shows that there exists $r \in \R_{>0}$ such that for all $t \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:

$(\text i) \quad f\left({z + t}\right) = f \left({z}\right) + t \left({f' \left({z}\right) + \epsilon \left({t}\right) }\right)$

where $B_r \left({0}\right)$ denotes an open ball of $0$, and $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \mathbb C$ is a continuous function with $\displaystyle \lim_{t \to 0} \ \epsilon \left({t}\right) = 0$.

Define $a, b, h, k \in \R$ by $a+ib = f' \left({z}\right)$, and $h+ik = t$.


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

\(\displaystyle u \left({x + h, y + k}\right)\) \(=\) \(\displaystyle \operatorname{Re}\left({f \left({z}\right) }\right) + \operatorname{Re}\left({tf' \left({z}\right) }\right) + \operatorname{Re}\left({t\epsilon \left({t}\right) }\right)\) $\quad$ Addition of Real and Imaginary Parts $\quad$
\(\displaystyle \) \(=\) \(\displaystyle u \left({x, y}\right) + \operatorname{Re}\left({ha - kb + ihb + iak}\right) + \operatorname{Re}\left({h \epsilon \left({t}\right) + ik \epsilon \left({t}\right) }\right)\) $\quad$ Definition of Complex Multiplication $\quad$
\(\displaystyle \) \(=\) \(\displaystyle u \left({x, y}\right) + ha - kb + h \operatorname{Re}\left({ \epsilon \left({t}\right) }\right) + k \operatorname{Re}\left({i \epsilon \left({t}\right) }\right)\) $\quad$ Multiplication of Real and Imaginary Parts $\quad$
\(\displaystyle \) \(=\) \(\displaystyle u \left({x, y}\right) + h \left({a + \operatorname{Re}\left({ \epsilon \left({t}\right) }\right) }\right) + k \left({-b + \operatorname{Re}\left({i \epsilon \left({t}\right) }\right) }\right)\) $\quad$ $\quad$

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:

$u \left({x + h, y}\right) = u \left({x, y}\right) + h \left({a + \operatorname{Re}\left({ \epsilon \left({h}\right) }\right) }\right)$

From Limits of Real and Imaginary Parts, it follows that $\displaystyle \lim_{h \to 0} \operatorname{Re}\left({ \epsilon \left({h}\right) }\right) = \operatorname{Re}\left({ \lim_{h \to 0} \epsilon \left({h}\right) }\right)= \operatorname{Re}\left({0}\right) = 0$.

From Continuity of Composite Mapping and Real and Imaginary Part Projections are Continuous, it follows that $\operatorname{Re}\left({ \epsilon \left({h}\right) }\right)$ is continuous as a function of $h$.

Then the Alternative Differentiability Condition proves that $\dfrac{\partial u}{\partial x} \left({x, y}\right) = a$.

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

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

$u \left({x, y + k}\right) = u \left({x, y}\right) + k \left({-b + \operatorname{Re}\left({i \epsilon \left({ik}\right) }\right) }\right)$

From Limits of Real and Imaginary Parts, it follows that $\displaystyle \lim_{k \to 0} \operatorname{Re}\left({i \epsilon \left({k}\right) }\right) = \operatorname{Re}\left({i \lim_{k \to 0} \epsilon \left({k}\right) }\right) = 0$.

Then the Alternative Differentiability Condition proves that $\dfrac{\partial u}{\partial y} \left({x, y}\right) = -b$.

$\Box$


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

\(\displaystyle v \left({x + h, y + k}\right)\) \(=\) \(\displaystyle v \left({x, y}\right) + \operatorname{Im}\left({ha - kb + ihb + iak}\right) + \operatorname{Im}\left({h \epsilon \left({t}\right) + ik \epsilon \left({t}\right) }\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle v \left({x, y}\right) + h \left({b + \operatorname{Im}\left({ \epsilon \left({t}\right) }\right) }\right) + k \left({a + \operatorname{Im}\left({i \epsilon \left({t}\right) }\right) }\right)\) $\quad$ by a similar calculation to the one above $\quad$

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:

$v \left({x + h, y}\right) = v \left({x, y}\right) + h \left({b + \operatorname{Im}\left({ \epsilon \left({h}\right) }\right) }\right)$

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

$\dfrac{\partial v}{\partial x} \left({x, y}\right) = b = - \dfrac{\partial u}{\partial y} \left({x, y}\right)$

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:

$v \left({x, y + k}\right) = v \left({x, y}\right) + k \left({a + \operatorname{Im}\left({i \epsilon \left({ik}\right) }\right) }\right)$

Here, the Alternative Differentiability Condition shows that:

$\dfrac{\partial v}{\partial y} \left({x, y}\right) = a = \dfrac{\partial u}{\partial x} \left({x, y}\right)$

This proves the first Cauchy-Riemann equation.

$\Box$

From Complex-Differentiable Function is Analytic, it follows that $f'$ is continuous.

From Continuity of Composite Mapping and Real and Imaginary Part Projections are Continuous, it follows that $\operatorname{Re} \left({f}\right)$ and $\operatorname{Im} \left({f}\right)$ are continuous.

Then all four partial derivatives are continuous, as

$\dfrac{\partial u}{\partial x} \left({x, y}\right) = \dfrac{\partial v}{\partial y} \left({x, y}\right) = a = \operatorname{Re} \left({f \left({z}\right) }\right)$
$\dfrac{\partial u}{\partial y} \left({x, y}\right) = - \dfrac{\partial v}{\partial x} \left({x, y}\right) = -b = -\operatorname{Im} \left({f \left({z}\right) }\right)$

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 $h, k \in \R \setminus \left\{ {0}\right\}$, and put $t = h + i k \in \C$.

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

Put:

$a = \dfrac{\partial u}{\partial x} \left({x, y}\right) = \dfrac{\partial v}{\partial y} \left({x, y}\right)$

and:

$b = - \dfrac{\partial u}{\partial y} \left({x, y}\right) = \dfrac{\partial v}{\partial x} \left({x, y}\right)$

From the Alternative Differentiability Condition, it follows that:

$u \left({x + h, y}\right) = u \left({x, y}\right) + h \left({a + \epsilon_{u x} \left({h}\right) }\right)$
$u \left({x, y + k}\right) = u \left({x, y}\right) + k \left({-b + \epsilon_{u y} \left({k}\right) }\right)$
$v \left({x + h, y}\right) = v \left({x, y}\right) + h \left({b + \epsilon_{v x} \left({k}\right) }\right)$
$v \left({x, y + k}\right) = v \left({x, y}\right) + k \left({a + \epsilon_{v y} \left({h}\right) }\right)$

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:

\(\displaystyle f \left({z + t}\right)\) \(=\) \(\displaystyle u \left({x + h, y + k}\right) + i v \left({x + h, y + k}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle u \left({x, y}\right) + iv \left({x, y}\right) + h \left({a + i b + \epsilon_{u x} \left({h}\right) + i \epsilon_{v x} \left({h}\right)}\right) + k \left({i a - b + \epsilon_{u y} \left({k}\right) + i \epsilon_{v y} \left({k}\right)}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle f \left({z}\right) + t \left({a + i b + \dfrac h t \epsilon_{u x} \left({h}\right) + \dfrac h t i \epsilon_{v x} \left({h}\right) - \dfrac k t i \epsilon_{u y} \left({k}\right) + \dfrac k t \epsilon_{v y} \left({k}\right)}\right)\) $\quad$ $\quad$

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:

\(\displaystyle \epsilon \left({t}\right)\) \(=\) \(\displaystyle \dfrac h t \epsilon_{u x} \left({h}\right) + \dfrac h t i \epsilon_{v x} \left({h}\right) - \dfrac k t i \epsilon_{u y} \left({k}\right) + \dfrac k t \epsilon_{v y} \left({k}\right)\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{\epsilon \left({t}\right) }\right\vert\) \(\le\) \(\displaystyle \left\vert{\dfrac h t \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t \epsilon_{v x} \left({h}\right) }\right\vert + \left\vert{\dfrac k t \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t \epsilon_{v y} \left({k}\right) }\right\vert\) $\quad$ Triangle Inequality for Complex Numbers $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{v x} \left({h}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{v y} \left({k}\right) }\right\vert\) $\quad$ Complex Modulus of Product of Complex Numbers $\quad$


Thus:

\(\displaystyle \lim_{t \mathop \to 0} \left\vert{\epsilon \left({t}\right) }\right\vert\) \(\le\) \(\displaystyle \lim_{t \mathop \to 0} \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{v x} \left({h}\right) }\right\vert\) $\quad$ Combined Sum Rule for Limits of Functions $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{v y} \left({k}\right) }\right\vert\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \lim_{t \mathop \to 0} \left\vert{ \epsilon_{ux} \left({h}\right) }\right\vert + \lim_{t \mathop \to 0} \left\vert{ \epsilon_{vx} \left({h}\right) }\right\vert\) $\quad$ as $\left\vert{\dfrac h t}\right\vert \le 1$ and $\left\vert{\dfrac k t}\right\vert \le 1$ by Modulus Larger than Real Part and Imaginary Part $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \lim_{t \mathop \to 0} \left\vert{ \epsilon_{uy} \left({k}\right) }\right\vert + \lim_{t \mathop \to 0} \left\vert{ \epsilon_{vy} \left({k}\right) }\right\vert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0\) $\quad$ $\quad$



This shows that:

$\displaystyle \lim_{t \mathop \to 0} \epsilon \left({t}\right) = 0$

From Continuity of Composite Mapping, it follows that $\epsilon \left({t}\right)$ is continuous.

Then the Alternative Differentiability Condition shows that:

$f' \left({z}\right) = a + i b$

$\Box$


Expression of Derivative

Let $z = x + i y$.

Then:

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

Similarly:

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

$\blacksquare$


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