# Real- and Complex- Analytic Functions.

1. Jul 23, 2010

### Bacle

Hi Again:

I don't know if this is obvious or not: An analytic complex function
f(z)=u(x,y)+iv(x,y) , can be made into an analytic function
f: R^2 -->R^2, since each of u(x,y) and v(x,y) is itself a real-analytic
function, i.e., we can use a standard argument by component function.

How about in the opposite direction, i.e., we have a real-analytic
function f(x), analytic in an interval (a,b). When can we extend
f(x) into a complex-analytic function.?. I suspect , thinking of power series,
that we can use the radius of convergence to construct an analytic function, i.e.,
if f(x) is analytic in (a-r,a+r), then f(x) can be extended to a complex-analytic
function in |z-a|<r .

Is this correct.?

2. Jul 24, 2010

### Anthony

This can be seen as follows: if $$f(z)=u(x,y)+\mathrm{i} v(x,y)$$ is complex analytic, then $$u$$ and $$v$$ are harmonic hence (real) analytic.

Yes, given a real analytic function on an (a,b), it can be extended to a complex analytic function.

3. Jul 24, 2010

### Petek

While Anthony's statement is true, the answer to Bacle's question is "no". Consider

$$f(x) = \frac{1}{1 + x^2}$$

in the interval (-1, 3), Using Bacle's notation, a = 1, r = 2. Then f is real-analytic in the interval (-1, 3), but f is not complex-analytic in |z - 1| < 2 due to the poles at +-i. The function f would be complex-analytic in some disk that didn't include the poles, however.

4. Jul 24, 2010

### Anthony

Apologies, I only read the first bit! That will teach me.

5. Jul 29, 2010

### Bacle

PeteK, Anthony: Thnaks for your posts. Please check on this; I would appreciate

Just a question: I guess we could look at 1/(1+x^2) on the real line as agreeing

with 1-x^2+x^4 -... =Sum_(x=0...n) x^2n on (-1,1). We can also see it as

the restriction of 1/(1+z^2) , to the real axis, so that, by the identity theorem

(using the fact that (-1,1) has limit points) , 1/(1+z^2) is the analytic extension

of ( 1/(1+x^2)) to the complex plane, and :

1/(1+z^2) = 1-z^2+z^4 -...... on |z|=1

But then it seems like 1/(1+x^2) on (-1,1) can be extended to the disk |z|=1.

And it seems like the above could be generalized to analytic functions defined on

the real line; by the comparison criterion, if f(x) were analytic in (-r,r) (we can

translate the function if necessary so that it converges in that interval) , then

we would have -1< |(a_nx^n)/ (a_(n-1)x^n-1)|<1 in (-r,r), which I believe

would imply that the analytic extension of f(x) to the complex plane would converge

in that same interval.?
real line

6. Jul 29, 2010

### Bacle

Sorry, I forgot to ask another followup:

Anthony: I understand that if f(z)=u+iv , then both u,v are harmonic, but
how does it follow from u,v being harmonic, that each is real-analytic.?.

I understand that from harmonicity, the second derivatives would be killed,
but it seems many other mixed partials would not be killed by u (equiv. v)
being harmonic.

Thanks.

7. Jul 29, 2010

### Dickfore

What's a real analytic function?

8. Jul 29, 2010

### Petek

First, let's consider Anthony's statement that

It would be more precise to say that "Any real analytic function on some open set on the real line can be extended to a complex analytic function on some open set of the complex plane." The function $f(x) = 1/(1 + x^2)$ shows that a real-valued function can be analytic on all of R, but not have a continuation that is analytic on all of C.

Next, in your above statement you probably mean the disk {z: |z| < 1} since the complex power series $1 - x^2 + x^4 - ...$ doesn't converge, for example, if z = 1.

Finally, in your last paragraph, do you mean something like "If f(x) is analytic in (r,r), then it has a continuation to the complex numbers that is analytic in the disk {z: |z| < r}"? If so, it not true as shown by the example of f(x) as given above and with any r > 1. If you meant something else, please restate.

9. Jul 29, 2010

### Petek

See the Wikipedia article on analytic functions.

10. Jul 29, 2010

### Bacle

Thanks, PeteK.

Would you please give a hint for why 1/(1+x^2) is real-analytic in
the interval (-1,3).? I can't see it. I can see how it agrees with:

Sum_i=0...oo x^(2n) =1-x^2+x^4-........ in (-1,1)

But I don't see how it is real-analytic beyond (-1,1).

P.S: you were right, the radius of convergence of f(z)=1-z^2+z^4-....

is 1 , and, yes, there are poles at +i, -i' my bad. Tho I think

f(z) can be continued beyond the disk.

Thanks.

11. Jul 29, 2010

### Hurkyl

Staff Emeritus
If you wanted to see whether or not it's real analytic at, say, 2, then you should look to see whether or not there is a power series defined on an interval containing 2.

12. Jul 29, 2010

### Bacle

"If you wanted to see whether or not it's real analytic at, say, 2, then you should look to see whether or not there is a power series defined on an interval containing 2. "

Yes, but that is precisely where I am stuck; I can see how :

1-x^2+x^4-x^6.....

is defined on (-1,1), and it agrees with 1/(1+x^2) , but I cannot

think of any other series defined outside of (-1,1), which agrees

with 1/(1+x^2). Can you spare a paradigm.?

13. Jul 29, 2010

### Petek

Do you know Taylor's Theorem and Taylor Series?

14. Jul 29, 2010

### Bacle

Yes, I do know both, but I have not seen them for a few years now;
I don't know if it is more accurate to say that I knew them.

Since the functions we are working with are real C<sup>oo</sup> ,
I guess the part of the theorem that you may be referring to is that,

I do know, e.g., that, in the real case, if the radius of convergence
of the Taylor series T(f(x)) of f(x) ) is r (let's standardize to (-r,r)
again, by translation) , then T(f(x)) does not converge to f(x)
either at r, or at -r , or both. Same goes for the complex case; there
must be a point of non-convergence in the boundary of the circle of
convergence, like in the case of 1/(1+z^2) , where you pointed out
that there are poles at +i and at -i . (tho we may be able to continue
f(z) analytically in C-{i,-i} . )

This is the property I was using for 1/(1+x^2) , which has radius of
convergence one, i.e., it converges in (-1,1), and diverges at x=1 and
at x=-1 .

15. Jul 29, 2010

### Hurkyl

Staff Emeritus
The point is that a disk centered on 0 is probably not the best choice if you're looking for a disk that contains 2.