Proving Domain D in C is Simply Connected

In summary: D can be continuously shrunk to a single point without leaving the domain. This proves that a domain D is simply connected if and only if, for every function f which is analytic and free of zeroes in D, a branch of the square root of f exists in D.In summary, a domain D is simply connected if and only if, for every function f which is analytic and free of zeroes in D, a branch of the square root of f exists in D. The existence of this branch of the square root also implies the existence of p^{th} roots of f in D.
  • #1
tarheelborn
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Homework Statement


Demonstrate that a domain [itex]D\in\mathbb{C}[/itex] is simply connected if and only if, for every function [itex]f[/itex] which is analytic and free of zeroes in [itex]D[/itex], a branch of the square root of [itex]f[/itex] exists in [itex]D[/itex].


Homework Equations





The Attempt at a Solution



I know that by definition, every function f which is analytic and free of zeroes in D, a branch of log(f(z)) exists in D and that this implies the existence of [itex]p^{th}[/itex] roots of that function. I am not sure how to get from here to square roots, though.
 
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  • #2
Can someone help me with the steps to prove this statement?

First, let's define a simply connected domain D as a domain in the complex plane where any closed curve can be continuously shrunk to a single point without leaving the domain. This implies that there are no holes or gaps in the domain.

Now, let's assume that for every function f which is analytic and free of zeroes in D, a branch of the square root of f exists in D. This means that for every point z in D, there exists a unique square root of f(z) in D.

To prove that D is simply connected, we need to show that any closed curve in D can be continuously shrunk to a single point without leaving the domain. So, let's take a closed curve C in D and a point z_0 in the interior of C.

Since f is analytic and free of zeroes in D, it can be expressed as f(z) = e^g(z), where g(z) is analytic in D. Now, using the branch of the square root of f in D, we can write f(z) = (e^g(z))^(1/2) = e^(g(z)/2). This implies that g(z)/2 is also analytic in D.

Now, let's define a new function h(z) = e^(g(z)/2), which is also analytic in D. Since g(z)/2 is analytic in D, it has a primitive function in D. This means that h(z) has a primitive function in D as well.

Since D is simply connected, we can use the Cauchy-Goursat theorem to evaluate the integral of h(z) along any closed curve C in D. This gives us:

∫h(z)dz = 0

But since h(z) is the square root of f(z), we can write this integral as:

∫h(z)dz = ∫(e^(g(z)/2))dz = ∫f(z)^(1/2)dz

Since f(z) is analytic and free of zeroes in D, the integral of f(z)^(1/2) along C is also equal to 0. This means that the integral of f(z)^(1/2) along any closed curve in D is equal to 0, which implies that f(z)^(1/2) is analytic and has a primitive function in D.

Therefore, D is simply
 

1. What is the definition of a simply connected domain?

A simply connected domain is a region in a two-dimensional space that is connected and has no holes or gaps. This means that any closed curve within the domain can be continuously deformed into a single point without leaving the domain.

2. How do you prove that a domain D is simply connected?

To prove that a domain D is simply connected, we need to show that any closed curve in D can be continuously deformed into a single point. This can be done by showing that any two points in D can be connected by a path that lies entirely within D, and that this path can be continuously deformed into a point without leaving D.

3. What is the importance of proving that a domain D is simply connected?

Proving that a domain D is simply connected is important in many areas of mathematics and physics. It allows us to apply certain theorems and techniques that are only valid for simply connected domains, such as the Cauchy integral theorem and the Poincaré lemma.

4. Can a domain D be both connected and simply connected?

Yes, a domain D can be both connected and simply connected. Being connected means that there are no holes or gaps in the domain, while being simply connected means that any closed curve in the domain can be continuously deformed into a single point. Both of these properties are satisfied by a disk or a rectangle, for example.

5. How can we prove that a domain D in C is simply connected using homotopy?

A common method for proving that a domain D in C is simply connected is by using homotopy. This involves showing that any two paths in D that have the same starting and ending points can be continuously deformed into each other without leaving D. If this is true for any two paths in D, then we can conclude that D is simply connected.

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