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 necessarily sufficient condition for a function to be holomorphic (it is a sufficient condition if, for example, the functions u and v have continuous partial derivatives). This system of equations was first published in 1814 by Cauchy, in his paper Sur les intégrales définies.
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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:
- <math>{ \partial u \over \partial x } = { \partial v \over \partial y}</math>
and
- <math>{ \partial u \over \partial y } = -{ \partial v \over \partial x}.</math>
The equations correspond structurally to the condition that the Jacobian matrix is of the form
- <math>
\begin{pmatrix}
a & -b \\ b & \;\; a
\end{pmatrix}, </math>
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.
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 <math>f(x + iy) = u(x, y) + iv(x, y)</math>, then:
- <math>{\partial u \over \partial x} = {\partial v \over \partial y}</math> and <math>{\partial v \over \partial x} = - {\partial u \over \partial y}</math>.
Now suppose <math>\bar f</math> is also analytic on D. Then since <math>f(x + iy) = u(x, y) - iv(x, y)</math>,
- <math>{\partial u \over \partial x} = -{\partial v \over \partial y}</math> and <math>{\partial v \over \partial x} = {\partial u \over \partial y}</math>.
Combining this with the early equations, we get:
- <math>{\partial u \over \partial x} = {\partial u \over \partial y} = {\partial v \over \partial x} ={\partial v \over \partial y} = 0</math>.
This shows that f is locally constant on D, and constant if D is connected.
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:
<math>f'(z)\,</math> <math>=\lim_{h\rightarrow 0} {f(z+h)-f(z) \over h}</math> <math>=\lim_{h\rightarrow 0}{u(x+h,y)+iv(x+h,y)-[u(x,y)+iv(x,y)]\over h}</math> <math>=\lim_{h\rightarrow 0}{[u(x+h,y)-u(x,y)]+i[v(x+h,y)-v(x,y)]\over h}</math> <math>=\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]}.</math>
This is now in the form of two difference quotients, so now
- <math>f'(z)={\partial u \over \partial x} + i {\partial v \over \partial x}.</math>
Taking the second path:
<math>f'(z)\,</math> <math>=\lim_{h\rightarrow 0} {f(z+ih)-f(z) \over ih}</math> <math>=\lim_{h\rightarrow 0}{u(x,y+h)+iv(x,y+h)-[u(x,y)+iv(x,y)]\over ih}</math> <math>=\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]}</math> <math>=\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]}</math> <math>=\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]}.</math>
Again, this is now in the form of two difference quotients, so
- <math>f'(z)={\partial v \over \partial y} - i {\partial u \over \partial y}.</math>
Equating these two we get
- <math>{\partial u \over \partial x} + i {\partial v \over \partial x} = {\partial v \over \partial y} - i {\partial u \over \partial y}.</math>
Equating real and imaginary parts, then
- <math>{\partial u \over \partial x} = {\partial v \over \partial y}</math>
- <math>{\partial u \over \partial y} = - {\partial v \over \partial x}. \quad\square</math>
Alternative formulation
Suppose <math>z = x + iy</math> for real variables x and y. Then we can write <math>x = (z + \bar z)/2</math> and <math>y = (z - \bar z)/(2i)</math>. Now <math>\mathit{x}</math> and <math>\mathit{y}</math> can be thought of as real functions of complex independent variables <math>\mathit{z}</math> and <math>\bar z</math>. Differentiating <math>\mathit{x}</math> and <math>\mathit{y}</math> gives:
- <math>{\partial x \over \partial z} = {1 \over 2}\ \mathrm{and}\ {\partial y \over \partial z} = {1 \over 2i}</math>
as well as
- <math>{\partial x \over \partial \bar z} = {1 \over 2}\ \mathrm{and}\ {\partial y \over \partial \bar z} = -{1 \over 2i}.</math>
Differentiate a function <math>f (x, y) = u(x, y)+iv(x, y)</math>:
- <math>{\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}.</math>
Finally, substitution yields:
- <math>{\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).</math>
If we let <math>{\partial f \over \partial \bar z} = 0</math>, then, since <math>{\partial f \over \partial x} = {\partial u \over \partial x} + i{\partial v \over \partial x}\ \mathrm{and}\ {\partial f \over \partial y} = {\partial u \over \partial y} + i{\partial v \over \partial y}</math>,
- <math>{\partial u \over \partial x} + i{\partial v \over \partial x} = {1 \over i}\left({\partial u \over \partial y} + i{\partial v \over \partial y}\right).</math>
This represents Cauchy-Riemann equations.
The relation has this interpretation: <math>\mathit{x}</math> and <math>\mathit{y}</math> must be constant with respect to <math>\bar z</math>. This expresses the concept that an analytic function is "truly" a function of a single complex variable, rather than of a real vector.
Polar representation
Considering the polar representation <math>z=re^{i\theta}</math>, the equations take the form
- <math>{ \partial u \over \partial r } = {1 \over r}{ \partial v \over \partial \theta},</math>
- <math>{ \partial v \over \partial r } = -{1 \over r}{ \partial u \over \partial \theta}.</math>
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
- <math>\bar{\partial}</math>
annihilates holomorphic functions. This generalizes most directly the formulation
- <math>{\partial f \over \partial \bar z} = 0</math>,
where
- <math>{\partial f \over \partial \bar z} = {1 \over 2}\left({\partial f \over \partial x} - {1 \over i}{\partial f \over \partial y}\right).</math>ca:Equacions de Cauchy-Riemann
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