- #1

lonewolf5999

- 35

- 0

Let f and g be analytic functions on a simply connected domain Ω such that f

^{2}(z) + g

^{2}(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:

f

^{2}+ g

^{2}= 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!