The Cauchy-Riemann equations

In order for a complex function of a  single complex variable to be differentiable it must be differentiable both parallell to the imaginary axis δy →0 and parallell to the real axis δx →0.

This condition leads to the CAuchy –Riemann equations-

The Cauchy–Riemann equations on a pair of real-valued functions of two real variables u(x,y) and v(x,y) are the two equations:

(1a)     \dfrac{ \partial u }{ \partial x } = \dfrac{ \partial v }{ \partial y } \,


(1b)    \dfrac{ \partial u }{ \partial y } = -\dfrac{ \partial v }{ \partial x } \,
Suppose that

 f(z) = u(z) + i \cdot v(z)

is a function of a complex number z. Then the complex derivative of ƒ at a point z0 is defined by

\lim_{\underset{h\in\mathbb{C}}{h\to 0}} \frac{f(z_0+h)-f(z_0)}{h} = f'(z_0)

provided this limit exists.

If this limit exists, then it may be computed by taking the limit as h → 0 along the real axis or imaginary axis; in either case it should give the same result. Approaching along the real axis, one finds

\lim_{\underset{h\in\mathbb{R}}{h\to 0}} \frac{f(z_0+h)-f(z_0)}{h} = \frac{\partial f}{\partial x}(z_0).

On the other hand, approaching along the imaginary axis,

\lim_{\underset{h\in \mathbb{R}}{h\to 0}} \frac{f(z_0+ih)-f(z_0)}{ih} =\frac{1}{i}\frac{\partial f}{\partial y}(z_0).

The equality of the derivative of ƒ taken along the two axes is

i\frac{\partial f}{\partial x}(z_0)=\frac{\partial f}{\partial y}(z_0),
Holomorphy is the property of a complex function of being differentiable at every point of an open and connected subset of \mathbb{C} (this is called a domain in \mathbb{C}). 
 function that is complex-differentiable in a whole domain (holomorphic) is the same as an analytic function. This is not true for real differentiable functions.

Om mattelararen

Licentiate of Philosophy in atomic Physics Master of Science in Physics
Detta inlägg publicerades i Advanced, Calculus, Imaginary numbers. Bokmärk permalänken.


Fyll i dina uppgifter nedan eller klicka på en ikon för att logga in:

Du kommenterar med ditt Logga ut /  Ändra )


Du kommenterar med ditt Twitter-konto. Logga ut /  Ändra )


Du kommenterar med ditt Facebook-konto. Logga ut /  Ändra )

Ansluter till %s