Is Cauchy Reimann condition sufficient for complex differentiability?

[kidsasd987]

Hi, I have a question about Cauchy Reimann equation

lets say
z=x+yi is in R^2
And there exists
f:R^2->R^2
f(z)=u(x,y)+v(x,y)i

Then cauchy reimann condition states that
If partial x of f and y of f are equal, then f is holomorphic

However, I am not sure how this can be a necesary sufficient condition because it only considers y and x derivatives but does not consider directional derivatives.

Because f (z) is essentially the same as a vector field, its component u (x,y) and v (x,y) will have directional derivatives at a given point p.

And by the definition of holomorphicity, it should have the same derivative regardless of direction of derivatives.

Could anyone explain me how the cuachy reimann condition implies this?

THANKS

 

[FactChecker]

The Cauchy-Riemann equations guarantee that the derivatives coming in from the real and imaginary directions are equal. That is enough to guarantee it from any direction.

 

[kidsasd987]

The Cauchy-Riemann equations guarantee that the derivatives coming in from the real and imaginary directions are equal. That is enough to guarantee it from any direction.

Could you please elaborate it further?

I searched up a little bit and found this,
http://math.stackexchange.com/quest...he-cauchy-riemann-equations-to-imply-differen

I get that n dimensional directional directional derivative can restore anyone specific directional derivative in R^n, but cannot fully understand the link between it and the equal derivatives for all direction( holomorphicity)

 

[mathwonk]

The complex numbers C are essentially R^2 equipped with a choice of 90 degree rotation, multiplication by i. The derivative of a map is a linear approximation to that map. so the real derivatives, ∂u/∂x, ∂u/∂y, ∂v/∂x, ∂v/∂y, form a matrix that gives the real linear approximation. we define a map C-->C to be holomorphic at a point if the linear approximation is also complex linear, which means it commutes with multiplication by i, or that choice of 90 degree rotation. If you apply that requirement to a matrix, that it commute with 90 degree counterclockwise rotation, you should see that it means the diagonal entries are equal, and the off diagonal entries are negatives of each other. these are the so called cauchy riemann conditions.

i.e. a map C-->C is real differentiable if it has a real linear approximation as a map R^2-->R^2. The map is also holomorphic if that real linear approximation is in fact complex linear, which means its matrix has the form given in the cauchy riemann equations. this is the way to understand it. the usual tricky derivation taking different directions for the complex limit has no motivation and gives no insight.

a fancier way to say it is to check that every real linear map can be decomposed into a complex linear part and a complex conjugate - linear part. then the cauchy riemann formula is the formula for the conjugate linear part, and the condition of holomorphicity is that this conjugate linear part should be zero.

in this decomposition the complex linear part is denoted by ∂f/∂z and the conjugate - linear part is denoted by ∂f/∂zbar. thus the c-r condition says equivalently that ∂f/∂zbar = 0.

 

[lavinia]

When the StackExchange post talks of directional derivatives it means limits of complex Newton quotients not directional derivatives of a function of two real variables.

So if ,for example, one takes the derivative in the purely imaginary direction one has the limit of the Newton quotient $lim_{h→0}(f(z + hi) - f(z))/hi$ not $lim_{h→0}(f(z + hi) - f(z))/h$

The complex derivative of a holomorphic function is the same in every direction. But the real directional derivative, the derivative of the vector field as you called it, is not.

For instance for the function $f(z)= z^2$,the complex derivative is $2z$. But the directional derivative of the vector field $(u^2-v^2,2uv)$ in the $y$ direction is $(-2v,2u)$ which is the 90 degree counterclockwise rotation of $2z$. That is: it is $2iz$.

On the other hand the complex directional derivative in the 1 direction is $lim_{h→0}((z+h)^2-z^2)/h$ and this is $2z$ again. It is also the directional derivative of the vector field. So in this case the real and complex directional derivatives are the same. Note that the complex derivative is the same in both directions but the real directional derivative is not.

As Mathwonk explained, the derivative of a holomorphic function is just multiplication by a complex number. This means that the partial derivatives of its real and complex parts satisfy the Cauchy-Riemann equations. Conversely, if the derivative of a map from $R^2$ into $R^2$ satisfies the Cauchy Riemann equations then it is multiplication by a complex number.

What you need to show is that if the derivative of a smooth map from $R^2$ to $R^2$ is multiplication by a complex number then the function is complex differentiable.