Cauchy-Riemann equations

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In mathematics, the Cauchy-Riemann differential equations in complex analysis, named after Augustin Cauchy and Bernhard Riemann, are two partial differential equations which provide a necessary but not sufficient condition for a function to be holomorphic. With additional conditions, such as the real and imaginary parts of the function -- the real functions u and v -- having continuous partial derivatives, satisfaction of the equations becomes equivalent to the analyticity of the complex function. This system of equations first appeared in the works of D'Alembert in 1752. Later, in 1777, Euler connected this system to the analytic functions. Cauchy used these equations to construct his theory of functions in 1814 (see his paper, Sur les intégrales définies). Riemann's dissertation on the theory of functions appeared in 1851.

1 Formulation
2 Example
3 Derivation
4 Alternative formulation
5 Polar representation
6 Several variables

[edit] Formulation

Let f(x + iy) = u + iv be a function from an open subset of the complex numbers C to C, where x, y, u, and v are real, and regard u and v as real-valued functions defined on an open subset of R². Then f is holomorphic if and only if u and v are continuously differentiable and their partial derivatives satisfy the Cauchy-Riemann equations, which are:

${ \partial u \over \partial x } = { \partial v \over \partial y }$

and

${ \partial u \over \partial y } = -{ \partial v \over \partial x } .$

There is a natural complex reformulation which provides immediate geometric insight:

${ i { \partial f \over \partial x } } = { \partial f \over \partial y } .$

The equations correspond structurally to the condition that the Jacobian matrix is of the form

$\begin{pmatrix} a & -b \\ b & \;\; a \end{pmatrix},$

the matrix representation of a complex number. Geometrically, this expresses the conformal nature by a combination of rotation and enlargement, for any analytic function at a point where its derivative isn't zero. It does so by a first-order picture (small discs are rotated and enlarged to other small approximate discs).

It follows from the equations, if they can be differentiated twice, that u and v must be harmonic functions since they satisfy Laplace's equation. The equations can therefore be seen as the conditions on a given pair of harmonic functions to come as real and imaginary parts of a complex-analytic function. For a given harmonic function u a corresponding harmonic function v is called a harmonic conjugate. If it exists it is unique up to a constant term.

[edit] Example

The equations give a direct insight into antiholomorphic functions.

Suppose a complex function f analytic on an open set D. Then f satisfies Cauchy-Riemann equations; that is, if $f (x + i y) = u (x, y) + i v (x, y)$ , then:

${\partial u \over \partial x} = {\partial v \over \partial y}$ and ${\partial v \over \partial x} = - {\partial u \over \partial y}$ .

Now suppose $\bar f$ is also analytic on D. Then since $\bar f(x + iy) = u(x, y) - iv(x, y)$ ,

${\partial u \over \partial x} = -{\partial v \over \partial y}$ and ${\partial v \over \partial x} = {\partial u \over \partial y}$ .

Combining this with the early equations, we get:

${\partial u \over \partial x} = {\partial u \over \partial y} = {\partial v \over \partial x} ={\partial v \over \partial y} = 0$ .

This shows that f is locally constant on D, and constant if D is connected.

[edit] Derivation

Consider a function f(z) = u(x, y) + i v(x, y) over C, and we wish to calculate its derivative at some point, z₀. We can essentially approach z₀ along the real axis towards 0, or down the imaginary axis towards 0.

If we take the first path:

$f'(z)\,$	$=\lim_{h\rightarrow 0} {f(z+h)-f(z) \over h}$
	$=\lim_{h\rightarrow 0}{u(x+h,y)+iv(x+h,y)-[u(x,y)+iv(x,y)]\over h}$
	$=\lim_{h\rightarrow 0}{[u(x+h,y)-u(x,y)]+i[v(x+h,y)-v(x,y)]\over h}$
	$=\lim_{h\rightarrow 0}{\left[\frac{u(x+h,y)-u(x,y)}{h}+i\frac{v(x+h,y)-v(x,y)}{h}\right]}.$

This is now in the form of two difference quotients, so now

$f'(z)={\partial u \over \partial x} + i {\partial v \over \partial x}.$

Taking the second path:

$f'(z)\,$	$=\lim_{h\rightarrow 0} {f(z+ih)-f(z) \over ih}$
	$=\lim_{h\rightarrow 0}{u(x,y+h)+iv(x,y+h)-[u(x,y)+iv(x,y)]\over ih}$
	$=\lim_{h\rightarrow 0}{\left[\frac{u(x,y+h)-u(x,y)}{ih} +i\frac{v(x,y+h)-v(x,y)}{ih}\right]}$
	$=\lim_{h\rightarrow 0}{\left[-i\frac{u(x,y+h)-u(x,y)}{h}+\frac{v(x,y+h)-v(x,y)}{h}\right]}$
	$=\lim_{h\rightarrow 0}{\left[\frac{v(x,y+h)-v(x,y)}{h}-i\frac{u(x,y+h)-u(x,y)}{h}\right]}.$

Again, this is now in the form of two difference quotients, so

$f'(z)={\partial v \over \partial y} - i {\partial u \over \partial y}.$

Equating these two we get

${\partial u \over \partial x} + i {\partial v \over \partial x} = {\partial v \over \partial y} - i {\partial u \over \partial y}.$

Equating real and imaginary parts, then

${\partial u \over \partial x} = {\partial v \over \partial y}$

${\partial u \over \partial y} = - {\partial v \over \partial x}. \quad\square$

[edit] Alternative formulation

Suppose $z = x + i y$ for real variables x and y. Then we can write $x = (z + \bar z)/2$ and $y = (z - \bar z)/(2i)$ . Now $x$ and $y$ can be thought of as real functions of complex independent variables $z$ and $\bar z$ . Differentiating $x$ and $y$ gives:

${\partial x \over \partial z} = {1 \over 2}\ \mathrm{and}\ {\partial y \over \partial z} = {1 \over 2i}$

as well as

${\partial x \over \partial \bar z} = {1 \over 2}\ \mathrm{and}\ {\partial y \over \partial \bar z} = -{1 \over 2i}.$

Differentiate a function $f (x, y) = u (x, y) + i v (x, y)$ :

${\partial f \over \partial z} = {\partial f \over \partial x}{\partial x \over \partial z} + {\partial f \over \partial y}{\partial y \over \partial z}\ \mathrm{and}\ {\partial f \over \partial \bar z} = {\partial f \over \partial x}{\partial x \over \partial \bar z} + {\partial f \over \partial y}{\partial y \over \partial \bar z}.$

Finally, substitution yields:

${\partial f \over \partial z} = {1 \over 2}\left({\partial f \over \partial x} + {1 \over i}{\partial f \over \partial y}\right)\ \mathrm{and}\ {\partial f \over \partial \bar z} = {1 \over 2}\left({\partial f \over \partial x} - {1 \over i}{\partial f \over \partial y}\right).$

If we let ${\partial f \over \partial \bar z} = 0$ , then ${\partial f \over \partial x} = -i {\partial f \over \partial y}$ and thus

${\partial u \over \partial x} + i{\partial v \over \partial x} = -i\left({\partial u \over \partial y} + i{\partial v \over \partial y}\right),$

which is equivalent to the Cauchy-Riemann equations.

[edit] Polar representation

Considering the polar representation $z = r e i θ$ , the equations take the form

${ \partial u \over \partial r } = {1 \over r}{ \partial v \over \partial \theta},$

${ \partial v \over \partial r } = -{1 \over r}{ \partial u \over \partial \theta}.$

Alternatively:

${\partial f \over \partial r} = {1 \over i r}{\partial f \over \partial \theta}$

where the derivatives evaluated on $r e i θ$ .

[edit] Several variables

There are Cauchy-Riemann equations, appropriately generalized, in the theory of several complex variables. They form a significant system of overdetermined PDEs. As often formulated, the d-bar operator

$\bar{\partial}$