# Complex analysis

1. Mar 9, 2012

### arthurhenry

There does not exist a non-constant analytic function in the unit circle which is real valued on the unit circle.

I am not able to see why. I am trying to apply Louisville's Theorem, or maybe Open Mapping Th., but I fail.
Is there a way of extending this function so that it entire? and even then...why should it be bounded?

Thank you

2. Mar 9, 2012

### mathwonk

well you can always count how many times a holomorphic function takes on a given value inside a circle by counting how many times it winds that circle around the given value.

if it is real valued on the unit circle then it does not wind around any point, so f(0) e.g,. would be taken on zero times, if f(0) is not real. But it is taken on at least once.

This is called the argument principle.

3. Mar 9, 2012

### arthurhenry

Thank you, this is a very good hint.Having said that this is where I am:

Each point p in unit disk D is mapped to a point. But (following your hint), f(p) is not assumen unless it is real, so f maps all of D to reals. Therefore, by Open Mapping Theorem, since the image has no interior points, f has to be constant.

I appreciate it if you could comment whether I got this or not.

4. Mar 9, 2012

### mathwonk

yeah it sounds roughly correct but maybe not quite precise. Your statement is true, but you want to prove it from stuff you know.

lets say it backwards.

lets see. by the open mapping theorem there is a point p in the disc such that f(p) is not real.

then since f(p) is not on the image of the unit circle, then the total order to which f assumes the value f(p) inside the disc equals the winding number about f(p) of the image by f of the unit circle and is at least one.

I.e. it equals the integral in the statement of the argument principle taken around the unit circle.

I guess that would be f'/[f - f(p)). Now you need to show that if the circle maps into the real axis, that you can choose a well defined branch of the log of that expression, defined on the whole unit circle,

and thus show that integral is zero, a contradiction.

5. Mar 13, 2012

### Bacle2

How about applying C_Riemann to the restriction of f to the boundary? (I'm a little confused; I usually understrand the circle to be S^1 , which has empty interior, so that it does not include the disk).

This restriction is also analytic, and of the form u_(x,y)+i .0 , i.e., v_(x,y)==0 .

Then C-R should do the rest.

6. Mar 16, 2012

### lavinia

The real and complex parts of the function are harmonic. if it is real on the boundary of the disc doesn't that mean that the complex part is zero on the boundary?

7. Mar 16, 2012

### Bacle2

It is tautologically true that if a complex-valued function is real-valued on the unit circle, then its imaginary part is zero.

8. Mar 17, 2012

### lavinia

that means the complex part must be constant in the interior since it is harmonic. Yes?

9. Mar 20, 2012

### Bacle2

Yes, by the maximum principle this would hold. Also, both u,v satisfy the MVProperty, and ,moreover, both are harmonic.

I don't see where you're getting at: do you disagree with any of the proposed solutions? By C-R restricted to the boundary, u is constant there ; v==0 , so use
u_x=0 , u_y=0. Then the complex part is 0 in the interior, so that, by the open-
mapping theorem, f must be constant. The other is the one worked out by th others.

10. Mar 20, 2012

### lavinia

No disagreement - just another look.

If the complex part is zero then Cauchy-Riemann tells you about the real part.

The open mapping theorem requires a fair amount of machinery. This way you just use definitions and simple properties of harmonic functions.

11. Mar 20, 2012

### Bacle2

O.K, no problem' always a good idea; I try to the extent possible to guide myself by "doubt everything that is not clear and distinct" . Still, as the complex part of f is ==0 in the interior--trivially harmonic-- I don't see how one could squeeze much out of it, but I have been wrong before--and surely will again.

Still, since the complex part is ==0 in the interior, one can repeat the argument that worked for the boundary.

And, re the OMTheorem, once you go over Rouche's theorem and the winding number, it seems kind of intuitive.