# Cauchy Riemann Equations

• Jan 2nd 2009, 07:33 PM
manjohn12
Cauchy Riemann Equations
In deriving the necessary conditions for a function $f(x) = u(x,y) + iv(x,y)$ to be differentiable at a point $z_0$, why do we get $f'(z_0) = v_{y}(x_{0}, y_{0}) - iu_{y}(x_{0}, y_{0})$? Would not it be $f'(z_0) = u_{y}(x_{0}, y_{0}) + iv_{y}(x_0, y_0)$?
• Jan 2nd 2009, 08:02 PM
Aryth
Because the Cauchy Riemann equations are:

$\color{red}\frac{\partial{u}}{\partial{x}} = \frac{\partial{v}}{\partial{y}}$

$\color{red}\frac{\partial{v}}{\partial{x}} = -\frac{\partial{u}}{\partial{y}}$

$i\frac{\partial{f}}{\partial{x}} = \frac{\partial{f}}{\partial{y}}$

$\frac{\partial{f}}{\partial{\bar{z}}} = 0$ where $\bar{z}$ is the complex conjugate of z = x + iy.

Where u and v are the real and imaginary parts of:

$f(x + iy) = u(x,y) + iv(x,y)$
• Jan 3rd 2009, 12:55 AM
Moo
Hello,
Quote:

Originally Posted by Aryth
Because the Cauchy Riemann equations are:

$\color{red}\frac{\partial{u}}{\partial{x}} = \frac{\partial{v}}{\partial{y}}$

$\color{red}\frac{\partial{v}}{\partial{x}} = \frac{\partial{u}}{\partial{y}}$

$i\frac{\partial{f}}{\partial{x}} = \frac{\partial{f}}{\partial{y}}$

$\frac{\partial{f}}{\partial{\bar{z}}} = 0$ where $\bar{z}$ is the complex conjugate of z = x + iy.

Where u and v are the real and imaginary parts of:

$f(x + iy) = u(x,y) + iv(x,y)$

No, the second one is $\frac{\partial v}{\partial x}={\color{red}-} \frac{\partial u}{\partial y}$
• Jan 3rd 2009, 06:34 AM
Aryth
You're right... I edited it.
• Jan 3rd 2009, 07:00 AM
HallsofIvy
I'm surprised that advertising like this would be allowed.
• Jan 3rd 2009, 07:33 AM
HallsofIvy
Quote:

Originally Posted by manjohn12
In deriving the necessary conditions for a function $f(x) = u(x,y) + iv(x,y)$ to be differentiable at a point $z_0$, why do we get $f'(z_0) = v_{y}(x_{0}, y_{0}) - iu_{y}(x_{0}, y_{0})$? Would not it be $f'(z_0) = u_{y}(x_{0}, y_{0}) + iv_{y}(x_0, y_0)$?

Fundamentally, the answer is the "chain rule". Since z= z+ iy, $\frac{dv}{dz}= \frac{\partial v}{\partial y}\frac{\partial y}{\partial z}$ and $\frac{\partial y}{\partial z}= \frac{1}{\frac{\partial z}{\partial y}}= \frac{1}{i}= -i$

Crucial point: 1/i= -i.

Another way of looking at the Cauchy-Riemann equations is this: With f(z)= f(x+ iy)= u(x,y)+ iv(x,y), the derivative, at $z= z_0$, is the given by the limit $\lim_{h\rightarrow 0} \frac{f(z_0+h)- f(z_0)}{h}$ just as for real numbers. And, just like limits for real numbers, in order that the limit exist, we must get the same result approaching $z_0$ along any path. In particular, if we approach $z_0$ along a line parallel to the real axis, h is real so we have
$\lim_{h\rightarrow 0}\frac{u(x_0+h,y_0)+ iv(x_0+ h,y_0)- u(x_0,y_0)- iv(x_0,v_0)}{h}$
$\lim_{h\rightarrow 0}\frac{u(x_0+h,y_0)- u(x_0,y_0)}{h}+ \lim_{h\rightarrow 0}\frac{iv(x_0+h,y_0)- iv(x_0,y_0)}{h}$
[tex]= \frac{\partial u}{\partial x}+ i\frac{\partial v}{\partial u}

Approaching instead along a line parallel to the imaginary axis, h is imaginary and we can use "ih", with i real, instead. Now we have
$\lim_{h\rightarrow 0}\frac{u(x_0,y_0+ h)+ iv(x_0,y_0+h)- u(x_0,y_0)- iv(x_0,y_0)}{ih}$
$\lim_{h\rightarrow 0}\frac{u(x_0, y_0+ h)- u(x_0, y_0)}{ih}+ \lim_{h\rightarrow 0}\frac{v(x_0,y_0+ h)- v(x_0,y_0)}{ih}$
and the important difference is that "i" in the denominator. It will cancel "i" in the v part and remember that 1/i= -i so
$= -i\lim_{h\rightarrow 0}\frac{u(u_0,y_0+h)- u(x_0,y_0)}{h}+ i\lim_{h\rightarrow 0}\frac{v(x_0,y_0+h)- v(x_0,y_0)}{h}$
$= -i\frac{\partial u}{\partial y}+ \frac{\partial v}{\partial y}$

In order that the limit exist, those two must be the same. Equating real and imaginary parts,
$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$ and
$\frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}$
• Jan 3rd 2009, 08:53 AM
Jester
Quote:

Originally Posted by HallsofIvy
I'm surprised that advertising like this would be allowed.

It's abuse of this wonderful site! Admins - you listening?