Analytic functions on simple connected region (complex analysis)

[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!

 

[Dick]

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?

 

[lonewolf5999]

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!

 

[Dick]

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.