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# Cauchy-Riemann equations

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 and sufficient condition for a function to be holomorphic.

Let f(x+iy) = u + iv be a function from an open subset of the complex numbers C to C, 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 differentiable and their partial derivatives satisfy the Cauchy-Riemann equations, which are:

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

and

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

It follows from the equations that u and v must be harmonic functions. 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.

## 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:

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

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:

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

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

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

## Polar representation

Considering the polar representation z = reiθ, the equations take the form

${ \partial u \over \partial r } = {1 \over r}{ \partial v \over \partial \theta},$
${ \partial v \over \partial r } = -{1 \over r}{ \partial u \over \partial \theta}.$
Last updated: 10-15-2005 01:52:04
03-10-2013 05:06:04