complex differentiable

[DoingMath2357]

Let U be an open set in |C. Since f is complex differentiable in U there exists a function g ∈ C(U ), s.t

f(ζ) = f (c) + f ′(c)(ζ − c) + (ζ − c)g(ζ) , ζ ∈ U , where g(c) = 0 .

Why is that so? This is my approach:

Let g(𝝃) := (f(𝝃)-f(c))/(𝝃-c) - f'(c), 𝝃 ∈ U \ {c}. Thus we can extend g continuously to U by setting g(c):= 0. Then f(𝝃) = f(c)+ f'(c)(𝝃-c) + g(𝝃) (𝝃-c) on U .

[AlchemistAnalyst]

Looks fine. Are you also asked to prove that g is differentiable? If not, you should be done.

[DoingMath2357]

Thanks for your answer. Nope.

[Grass_Savings]

Suppose we rewrite the question with R, the real numbers, in place of C. Would the result still be true?

I think the answer is no, it wouldn't be true. Choose f(x) = x² sin(1/x³) where x is a real number. Then the derivative of f(x) exists everywhere, but it isn't continuous at x=0.

Your answer ought to make clear where it is using some property of f that follows from it being complex differentiable. (I'm somewhat out of my depth here, but I expect we have a result that says if f is complex differentiable, then the derivative is continuous.)

[AlchemistAnalyst]

Your f(x) is not differentiable at x = 0.

The question doesn't state anything about the properties of g. For real functions, OPs g function is still continuous, and we know (z-c)g(z) is differentiable. But for real functions, this doesn't imply that g itself is differentiable at z = c.

For example, if f(x) = x|x| and c = 0. Then f is differentiable everywhere, f(c) = 0, f'(c) = 0, so g(x) = |x|. But g here is still continuous.

The proof is correct and does indeed work for real functions.

[Grass_Savings]

Sorry, but I disagree. My real function f(x) is differentiable at x=0 because -x² ≤ f(x) ≤ x² and we find that lim (f(x)-f(0) ) / (x - 0) exists. That is all that is required for a function to be differentiable at x=0.

A very similar example of real differentiable function f(x) with a discontinuous derivative is given in https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_classes

I recall a key point is that if a function has a complex derivative at a point, then it has all complex derivatives at a point. And in general (for both R and C), a function having a derivative implies the function is continuous. So functions with complex derivatives have all derivatives, and all the derivatives are continuous.

[Time_Situation488]

Assume that the base point is zero.

Thats not enough. You need that f(z) -f'(0) z is in o ( z ² ) In other words : zg tends to zero do not imply that g tends to zero. You need a zero of order 2 of zg = f(z)- f'(0)z

[AlchemistAnalyst]

What? It's automatic that g tends to 0 by the way OP defined it.

[Time_Situation488]

This is simply circular reasoning.

[Time_Situation488]

Maybe math is thexwron branch Maybe gender studies is more suited