# Homework Help: Complex derivative

1. Aug 21, 2011

### Andz001

I'm solving complex derivatives of holomorphic functions, my question is if I take the derivative of a holomorphic function will the result always be holomorphic too?

The examples I made always have a holomorphic result but I'm trying to find a counter example.

2. Aug 21, 2011

### spamiam

Welcome to PF!

How long have you been studying complex analysis? There is actually a major theorem on this subject whose content you can find on Wikipedia. A key part of the argument is the Cauchy integral formula, so you'll need some background in complex analysis to read the article.

3. Aug 21, 2011

### Andz001

Thanks for being the first person to help me :)
My experience with complex analysis is limited, but I know the general theorems.
So if f'(z)=$\partial$x f(z) and f'(z)= -i$\partial$y f(z). Then $\partial$x f(z) + i$\partial$y f(z) = 0. So the derivative of z which is holomorphic will always result in a holomporphic function. Is this correct?

4. Aug 21, 2011

### spamiam

Hm, I didn't understand your reply. Does $\partial_x$ mean $\frac{\partial}{\partial x}$? It looks like you're doing something like the proof of the Cauchy-Riemann equations.

Could you state your homework problem as it appears? Are you really to show that the derivative of a holomorphic function is holomorphic or to provide a counterexample?

5. Aug 22, 2011

### Andz001

The actual question is: If f(z) is holomorphic and continuously differentiable in an open area $\Omega$, then it's complex derivative f'(z) is holomorphic. True or False?

In my previous attempt ∂x means ∂/∂x.

6. Aug 22, 2011

### HallsofIvy

Have you looked at the definition of "holomorphic"? A function is holomorphic (on a set) if and only if it is analytic for all x in that set. When a set is not specified it is typically taken to be all complex numbers.

And, of course, one definition of "analytic" at a point is that its Taylor series exist at that point and converge to the value of the function in some neighborhood of the point. If a function is "holomorphic", it follows that its Taylor series exists and converge to the value of the function for all complex numbers.

7. Aug 22, 2011

### spamiam

I disagree. That is not the definition of holomorphic, but rather an important theorem in complex analysis. This is the definition that I'm familiar with: http://planetmath.org/encyclopedia/Holomorphic.html [Broken].

Are you supposed to provide a proof, Andz? If not, HallsofIvy's post contains the answer.

Last edited by a moderator: May 5, 2017