# Help with a unit disc property for a holomorphic function

1. Jan 30, 2012

### Stephen88

1. The problem statement, all variables and given/known data
Suppose that f is holomorphic in an open disc U and that Re(f) is
constant in U. I have to show that f must be constant in U. Also what is the essential
property of the disc U that it used here? Give an example of an open
set U for which the conclusion fails.

2. Relevant equations
Cauchy–Riemann equations.

3. The attempt at a solution
Let f=u+vi where u is a constant.Since f is holomorphic by the Cauchy–Riemann equations->
u_x=v_y and u_y=-v_x but since u is a constant u_x=u_y=0 => 0=v_y =-v_x...therefore f is constant.
The disc U has to be open,as in:U(a,r)={z:|z-a|<r}.
Is this correct?What should I do for the last part?
Thank you

2. Jan 30, 2012

### Stephen88

will U(0,1) be the special case?

3. Jan 30, 2012

### Dick

Suppose the open set is the union of two disks that don't intersect?

4. Jan 30, 2012

### Stephen88

I'm sorry but I don't know what to answer to that question.

5. Jan 30, 2012

### Dick

Any reason why f can't have one constant value on the first disk and a different constant value on the second disk?

6. Jan 30, 2012

### Stephen88

To be honest I don't think so...that's like having a large supermarket divided in two...one or more products may be on both sides.

7. Jan 30, 2012

### Dick

That's a strange analogy. Look. Define f(x)=1 on U(0,1) and f(x)=2 on U(3,1). Let the domain D be the union of the two disks. i) is there any place in the domain where the derivatives of f aren't zero? ii) is f constant on the domain?

8. Jan 30, 2012

### Stephen88

No there isn't...also I think that if there are 2 or more disjoint open sets that make U=> in a contradiction

9. Jan 30, 2012

### Dick

I hope that means you get it. So what's the 'essential property' of the domain that would eliminate these cases?

10. Jan 30, 2012

### Stephen88

U is connected

11. Jan 30, 2012

### Dick

That's it.

12. Jan 30, 2012

### Stephen88

Thank you:)

13. Jan 30, 2012

### jeniedieu

Doesn't such choice of f(x) contradict the given fact in this problem that "Re(f) is constant in U" ?

14. Jan 30, 2012

### Dick

Good point. I lost track of the problem. Make that f(x)=i and f(x)=2i.