Help with a unit disc property for a holomorphic function

In summary, the problem states that if f is holomorphic in an open disc U and Re(f) is constant in U, then f must also be constant in U. The essential property of U in this proof is that it must be connected. As for an example of an open set U where the conclusion fails, consider the union of two disjoint discs, such as U(0,1) and U(3,1), where f(x)=i on U(0,1) and f(x)=2i on U(3,1). This contradicts the given fact that Re(f) is constant in U.
  • #1
Stephen88
61
0

Homework Statement


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.

Homework Equations


Cauchy–Riemann equations.

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
 
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  • #2
will U(0,1) be the special case?
 
  • #3
Suppose the open set is the union of two disks that don't intersect?
 
  • #4
I'm sorry but I don't know what to answer to that question.
 
  • #5
Stephen88 said:
I'm sorry but I don't know what to answer to that question.

Any reason why f can't have one constant value on the first disk and a different constant value on the second disk?
 
  • #6
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
Stephen88 said:
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.

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
No there isn't...also I think that if there are 2 or more disjoint open sets that make U=> in a contradiction
 
  • #9
Stephen88 said:
No there isn't...also I think that if there are 2 or more disjoint open sets that make U=> in a contradiction

I hope that means you get it. So what's the 'essential property' of the domain that would eliminate these cases?
 
  • #10
U is connected
 
  • #11
Stephen88 said:
U is connected

That's it.
 
  • #12
Thank you:)
 
  • #13
Dick said:
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.

Doesn't such choice of f(x) contradict the given fact in this problem that "Re(f) is constant in U" ?
 
  • #14
jeniedieu said:
Doesn't such choice of f(x) contradict the given fact in this problem that "Re(f) is constant in U" ?

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

FAQ: Help with a unit disc property for a holomorphic function

What is a unit disc property for a holomorphic function?

A unit disc property for a holomorphic function refers to the property that a function is holomorphic on the unit disc in the complex plane. This means that the function is differentiable at every point within the unit disc and has a unique derivative at each point within the disc.

How is the unit disc property related to the Cauchy-Riemann equations?

The unit disc property is closely related to the Cauchy-Riemann equations, which are a set of necessary and sufficient conditions for a function to be holomorphic. The Cauchy-Riemann equations state that the partial derivatives of the function with respect to the real and imaginary components of the complex variable must satisfy certain conditions, which are also satisfied for functions with the unit disc property.

What is the significance of the unit disc property for holomorphic functions?

The unit disc property is significant because it allows for the use of powerful tools and techniques from complex analysis, such as the Cauchy integral formula and Laurent series, to study and solve problems involving holomorphic functions. It also has important applications in fields such as physics and engineering.

How can the unit disc property be used to determine the behavior of a holomorphic function?

The unit disc property can be used to determine the behavior of a holomorphic function by analyzing its singularities within the unit disc. If a function has poles or essential singularities within the disc, this can greatly affect its behavior and the techniques that can be used to study it.

Are there any known counterexamples to the unit disc property for holomorphic functions?

Yes, there are known counterexamples to the unit disc property for holomorphic functions. One example is the function f(z) = 1/z, which is holomorphic on the unit disc except at the origin, where it has a singularity. This function does not satisfy the unit disc property, as it is not differentiable at the origin.

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