CR Equations: Real & Imaginary Parts Satisfy Cont. & Diff.

[fleazo]

Complex differentiable <--> real and imaginary parts satisfy C-R eqns and are cont.

Say we have a complex function f(z) we can break this into real and imaginary parts:

f(z)=u(x,y)+iv(x,y)In my book I am told the following:(1) f complex differentiable at z₀ in ℂ --> the Cauchy Reimann equations are satisfied by the partials of u and v at z₀I am also told seperately:(2) For some z₀ in ℂ, if the partials of u and v exist and are continuous on an open set containing z₀ and they satisfy the Cauchy Reimann equatiions, then f is complex differentiable at z₀Why isn't this an iff thing? It seems like the only thing keeping it from being bidiriectional is the fact that in (2) I also have to check that the partials are continuous on an open set O containing z₀. But this is where my doubt arises. I am also told in my book:(3) If u and v satisfy the Cauchy Reimann equations at z₀ for some neighborhood O containingi z₀, then u and v are both harmonic in O. Since being harmonic requires establishing an equality between the second partial derivatives, isn't such a thing only possible if the first partial derivatives are continuous (as continuity is a essential for differentiability)? So how if u and v are harmonic shouldn't their first partials be continuous? If (1) gaurantees the Cauchy Reimann equations are satisfied which allows me to invoke (3), then doesn't (1) also gaurantee that the partials are continuous? So why can't I simply say f is differentiable at z₀ in ℂ ⇔ it's real and imaginary components have partial derivatives continuous on an open set O containing z₀ and these partials satisfy the Cauchy Reimann equations. Why does this need to be broken into two separate theorems?SIDE QUESTION: I know that if a complex function f is differentiable at z₀ it is actually infinitely differentiable there. But are it's real and imaginary parts also infinitely differentiable? I am just getting so confused separating what I can infer about the complex function and what I can infer about its real and imaginary parts which are functions of real variables.

 

[Erland]

To conclude that f(z) is infinitely complex differentiable at a point z0 and that its real and imaginary parts have continuous partial derivatives of all orders in a neighborhood of z0, it is not sufficient to assume that f(z) is complex differentiable at z0, but we need to assume that this complex differentiability hold for all points in a neighborhood of z0, which is the same as saying that f is analytic (or holomorphic) at z0.

In your proposed equivalence, you must therefore not talk only about a single point z0, but about all points in some region in C.

For example, if we put f(z)=z^2 if |z| is rational and 0 otherwise, then f'(0) exists and is 0 at 0, and the first order partial derivatives of the real and imaginary parts of f exist and satisfy the Cauchy-Riemann equtaions at 0 (they are all 0 there), but your proposed theorem fails there, for the partial derivatives do not exist at any point other than 0, and furthermore, f is not complex differentiable or even continuous at any other point and f is only complex differentiable one time at 0.
Your proposed theorem therefore fails for this f, but the original implication that complex differentiability at a point implies Cauchy-Riemann at that point is true also for this f.

 

[fleazo]

ok, I think I see where I was going wrong. So in (3) that I had posted originally, this is the condition where the partials satisfy the C-R equations throughout a neighborhood containing z₀, which means they are continuous in that neighborhood and f is holomorphic at z₀. So the equivalence holds if f is holomorphic at z₀, not simply complex differentiable.


FUrthermore, it seems you can have a function f that is complex differentiable at z₀, so the C-R equations are satisfied at z₀, but the partials are not necessarily harmonic. This happens when f is complex differentiable at z₀ but not holomorphic at z₀

 

[Bacle2]

ok, I think I see where I was going wrong. So in (3) that I had posted originally, this is the condition where the partials satisfy the C-R equations throughout a neighborhood containing z₀, which means they are continuous in that neighborhood and f is holomorphic at z₀. So the equivalence holds if f is holomorphic at z₀, not simply complex differentiable.


FUrthermore, it seems you can have a function f that is complex differentiable at z₀, so the C-R equations are satisfied at z₀, but the partials are not necessarily harmonic. This happens when f is complex differentiable at z₀ but not holomorphic at z₀

Take f(z)=|z|^2 = x^2+y^2 .

Then u_x=2x ; u_y=2y ; v_x=0=v_y , u_x(0)=v_y ; u_y(0)=v_x , and

u_xx=2 =u_yy.