(Complex) Analytic as opposed to Differentiable

1. Apr 22, 2010

phyzmatix

Good day!

Well, I've just started this year with Complex Analysis (we're using "Complex Variables and Applications" 8th ed by Brown and Churchill) and as I'm going through some assignment questions, I noticed that one of the questions states:

"Is g analytic at any point of C (as opposed to differentiable)? Justify your answer."

Does this make any sense?

My understanding of complex analytic functions is that they are functions that are everywhere differentiable so how can one have analyticity (is that actually a word?) as opposed to differentiablility seeing that (to my understanding) the two are inseparable?

I'd appreciate any further explanation/links (other than Wikipedia, I've been there thanks! ).

phyz

2. Apr 22, 2010

ice109

analytic means you can take infinitely many derivatives. differentiable means you can take 1

3. Apr 22, 2010

Landau

You have to be careful what definitions to use. I've read several definitions of "analytic", the one I learned is "locally having a power series". Then there's the notion of "holomorphic", which usually means "locally being (complex) differentiable". These notions are shown to be equivalent for complex functions.

[Sometimes people require the derivative to be continuous in the definition of holomorphic; eventually all notions are equivalent, but it can be confusing when reading different texts.]

4. Apr 23, 2010

phyzmatix

I'm sorry, but I don't get this. Infinitely many derivatives as in infinitely many higher order derivatives?

Ok, I hear you, but how would you interpret that question then? (on re-reading this question I realise it sounds a bit snotty, but it really isn't...limitations of communication through text huh? )

5. Apr 23, 2010

Landau

Probably they mean what ice109 says; and yes, with infinitely many derivatives he means infinitely many higher order derivatives. Take the derivative, the result is differentiable, take the derivative, the result is differentiable, take the derivative, etc. As opposed to: take the derivative, the result may not be differentiable ;)

6. Apr 23, 2010

phyzmatix

Got it! Thanks!

7. Apr 23, 2010

I thought differentiable functions on the complex plane were automatically infinitely differentiable?

8. Apr 23, 2010

fluidistic

If when you say "on the complex plane" you mean over the entire complex plane then the function is said be entire. In this case yes, it is infinitely many times derivable. Now if you restrict the analiticity in an interval... I'm not sure, I'd ask a mathematics student.

9. Apr 23, 2010

Wikipedia says:

"Holomorphic functions are complex functions defined on an open subset of the complex plane which are differentiable. Complex differentiability has much stronger consequences than usual (real) differentiability. For instance, holomorphic functions are infinitely differentiable, a fact that is far from true for real differentiable functions."

and there is a more detailed proof:

http://en.wikipedia.org/wiki/Proof_that_holomorphic_functions_are_analytic

that if a complex function is differentiable at every point within on open disk (i.e. holomorphic) then it is infinitely differentiable (analytic).

10. Apr 23, 2010

fluidistic

"Then it is infinitely differentiable (analytic)" ok but where? Only in this disk I guess. Otherwise any analytic function would be entire, which is not true in my opinion.

11. Apr 23, 2010

Yes I think that's right - analytic means its power series converges within that open disk. This means that if a function is differentiable in some open neighbourhood, it must also be infinitely differentiable in that neighbourhood.

12. Apr 24, 2010

Bacle

I thiknk the issue here is that Cauchy-Riemann must be satisfied in a whole 'hood
( 'hood =neighborhood *) of your point, and not just at a single point z0.
In the last case, f is just differentiable at z0 , while if C-R is satisfied
in a neighborhood of z0. This means that the differential quotient exists
(its limit exists) only at one point, but not in a 'hood of the point.

Only example I can think of now is that of |z|2 , the modulus function:

We have : |z|2= x2+y2

Then ux=2x+2y , and vy= 0

(maybe using a real-valued complex function is somewhat-contrived, and/or
"cheating" , but I think it somewhat makes the point.)

Above lines imply that C-R is satisfied only at the origin. You can check in more
detail by evaluating the quotient [f(z+h)-f(z)]/h , as |h|->0 , and check that
the limit exists only at the origin.

Now, it would be interesting to find a complex-valued function that is differentiable
but not analytic. I will think about it.

* I read this in someone else's post, using 'hood for neighborhood, I thought it
was funny.

13. Apr 24, 2010

Bacle

Re: (Complex) Analytic as opposed to Differentiable. Corrections

I think the issue here is that Cauchy-Riemann must be satisfied in a whole 'hood
( 'hood =neighborhood *) of your point, and not just at a single point z0.
In the last case, f is just differentiable at z0 , while if C-R is satisfied
in a neighborhood of z0, then f is analytic at z0.
This means that the differential quotient exists
(its limit exists) only at one point, but not in a 'hood of the point.

Only example I can think of now is that of |z|2 , the modulus function:

We have : |z|2= x2+y2

Then u(x,y)=x2+y2, and v(x,y)==0 ,

so that ux=2x , uy=2y

, and vx(x,y)=vy(x,y)= 0 , so that C-R is satisfied only

at the origin.

(maybe using a real-valued complex function is somewhat-contrived, and/or
"cheating" , but I think it somewhat makes the point.)

Above lines imply that C-R is satisfied only at the origin. You can check in more
detail by evaluating the quotient [f(z+h)-f(z)]/h , as |h|->0 , and check that
the limit exists only at the origin.

Now, it would be interesting to find a complex-valued function that is differentiable
but not analytic. I will think about it.

* I read this in someone else's post, using 'hood for neighborhood, I thought it
was funny.

14. Apr 24, 2010

Bacle

This depends on how your function is expressed: if it is expressed, e.g., as a power series,
then you can evaluate its radius of convergence.

Otherwise, set up C-Riemann. Wherever C-R is satisfied by f, f will have a power series
convergent to the value of f at each point.

15. Apr 25, 2010

Landau

In complex analysis we always work with open subsets $$U\subset\mathbb{C}$$. A function $$f:U\to \mathbb{C}$$ is analytic (meaning analytic at every point of U) if and only if it is holomorphic (meaning holomorphic, i.e. complex differentiable, at every point of U).

But we don't know if TS has already developed this important theorem; maybe the course just started and it is a question from one of the first chapters.