# Analytic functions on simple connected region (complex analysis)

• lonewolf5999
In summary: Since f+ig is continuous on Ω, it is also analytic on Ω. So, by the theorem, f+ig = exp(H), which gives you the results you were looking for!
lonewolf5999
Here's the problem:
Let f and g be analytic functions on a simply connected domain Ω such that f2(z) + g2(z) = 1 for all z in Ω. Show that there exists an analytic function h such that f(z) = cos (h(z)) and g(z) = sin(h(z)) for all z in Ω.

Here's my attempt at a solution:
f2 + g2 = 1 on Ω, so (f+ig)(f-ig) = 1 on Ω, so neither of those expressions are ever 0 on Ω. Then, defining H = (1/i)(f' + ig')/(f+ig), H is a composition of analytic functions, and hence is analytic on Ω, a simply connected domain. Therefore H has an antiderivative h on Ω.

Now I'd like to argue that cos(h(z)) = f(z) and similarly for g(z), but it's not clear to me how to do this. If we integrate H directly (I know this is not allowed, but supposing it is), we get something like h = (1/i) log(f+ig), so that exp(ih) = exp(log(f+ig)) = f+ig, but on the other hand exp(ih) = cos(h) + i sin(h), which is more or less what I want. I've checked that cos(h(z)) = f(z), the other equality follows, and conversely as well, so my problem is in showing that one of those equality holds without using the other.

I'd really appreciate any help on how to proceed from my definition of H to argue that its antiderivative h satisfies f(z) = cos (h(z)) and g(z) = sin(h(z)). Thanks in advance!

lonewolf5999 said:
Here's the problem:
Let f and g be analytic functions on a simply connected domain Ω such that f2(z) + g2(z) = 1 for all z in Ω. Show that there exists an analytic function h such that f(z) = cos (h(z)) and g(z) = sin(h(z)) for all z in Ω.

Here's my attempt at a solution:
f2 + g2 = 1 on Ω, so (f+ig)(f-ig) = 1 on Ω, so neither of those expressions are ever 0 on Ω. Then, defining H = (1/i)(f' + ig')/(f+ig), H is a composition of analytic functions, and hence is analytic on Ω, a simply connected domain. Therefore H has an antiderivative h on Ω.

Now I'd like to argue that cos(h(z)) = f(z) and similarly for g(z), but it's not clear to me how to do this. If we integrate H directly (I know this is not allowed, but supposing it is), we get something like h = (1/i) log(f+ig), so that exp(ih) = exp(log(f+ig)) = f+ig, but on the other hand exp(ih) = cos(h) + i sin(h), which is more or less what I want. I've checked that cos(h(z)) = f(z), the other equality follows, and conversely as well, so my problem is in showing that one of those equality holds without using the other.

I'd really appreciate any help on how to proceed from my definition of H to argue that its antiderivative h satisfies f(z) = cos (h(z)) and g(z) = sin(h(z)). Thanks in advance!

There is a theorem that if a(z) is analytic and nonzero in a simply connected domain then there is an analytic function b(z) such that a(z)=exp(b(z)). That's what you are looking for isn't it?

I can't find that theorem in my book, but if I use it, the answer comes out very easily: since f+ig is analytic and non-zero on Ω, let H be an analytic function such that f+ig = exp(H), then define h = iH, and I can show that f = cos(h), g = sin(h). I guess I'll just prove it for my problem. Thanks for the help!

lonewolf5999 said:
I can't find that theorem in my book, but if I use it, the answer comes out very easily: since f+ig is analytic and non-zero on Ω, let H be an analytic function such that f+ig = exp(H), then define h = iH, and I can show that f = cos(h), g = sin(h). I guess I'll just prove it for my problem. Thanks for the help!

It's really pretty easy to see if you know how the log function works. As long as your domain stays way from zero and doesn't contain any loops around zero, you can define a continuous log function on the domain.

## 1. What is a simple connected region in complex analysis?

A simple connected region in complex analysis is a subset of the complex plane where any two points can be connected by a continuous path without leaving the region. This means that the region has no holes or gaps, and is simply connected.

## 2. What are analytic functions in complex analysis?

Analytic functions are functions that are differentiable at every point in a given region of the complex plane. This means that they have a derivative at every point in their domain and can be represented by a power series.

## 3. What is the Cauchy-Riemann equation and how is it related to analytic functions?

The Cauchy-Riemann equation is a set of conditions that must be satisfied for a function to be analytic. It relates the partial derivatives of a function's real and imaginary parts, and if these conditions are met, the function is analytic in the given region.

## 4. How are contour integrals used in the study of analytic functions?

Contour integrals are used to evaluate the values of analytic functions along a given curve in the complex plane. They can also be used to calculate the values of derivatives and integrals of these functions.

## 5. What are some applications of analytic functions in real-world problems?

Analytic functions have many practical applications, such as in fluid dynamics, electromagnetism, and signal processing. They are also used in the study of complex numbers and in the development of mathematical models for various physical phenomena.

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