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I am studying for a qualifier exam in complex analysis and right now I'm solving questions from old exams. I am trying to prove the following:

Prove that if $f$ and $g$ are entire functions such that $f(z)^2 + g(z)^2 = 1$ for all $z \in \mathbb{C}$, then there exists an entire function $h$ such that $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$.

My Attempt

The approach that occurred to me is the following. Since $f(z)^2 + g(z)^2 = 1$ then we have $(f(z) + ig(z))(f(z) - ig(z)) = 1$. Then each factor is nonvanishing everywhere in $\mathbb{C}$ and thus by the "holomorphic logarithm theorem" we know that since $\mathbb{C}$ is simply connected, there exists a holomorphic function $H:\mathbb{C} \to \mathbb{C}$ such that

$$e^{H(z)} = f(z) + ig(z)$$

and then we can write $\exp(H(z)) = \exp\left(i\dfrac{H(z)}{i} \right) = \exp(ih(z))$,

where $h(z) := \dfrac{H(z)}{i}$.

Thus so far we have an entire function $h(z)$ that satisfies

$$e^{ih(z)} = f(z) + ig(z)$$

On the other hand, we also know that $e^{iz} = \cos{z} + i \sin{z}$ for any $z \in \mathbb{C}$, thus we see that

$$e^{ih(z)} = \cos{(h(z))} + i \sin{(h(z))} = f(z) + ig(z)$$

Thus at this point I would like to conclude somehow that we must have $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$, but I can't see how and if this is possible.

My questions

  1. Is the approach I have outlined a correct way to proceed, and if so how can I finish my argument?
  2. If my argument does not work, how can this be proved?

Thanks for any help.

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Since $f-ig$ is the reciprocal of $f+ig$, you know that $e^{-ih(z)}=f(z)-ig(z)$. Average this with $e^{ih(z)}$ to finish. (I am not familiar with the holomorphic logarithm theorem you mention though.) – anon Jul 6 '12 at 4:21
@anon Maybe that name is not the usual one. I have only seen it by that name in Greene and Krantz's book Function Theory of One Complex Variable. It is stated as Lemma 6.6.4 in page 197 of the book. – Adrián Barquero Jul 6 '12 at 4:25
Just in case the page view from google books doesn't work, it basically says that if $U$ is a simply connected open set and if $f: U \to \mathbb{C}$ is holomorphic and nowhere zero on $U$, then there exists a holomorphic function $h$ on $U$ such that $e^h \equiv f$ on $U$. – Adrián Barquero Jul 6 '12 at 4:28
Ah, of course. ${}$ – anon Jul 6 '12 at 4:32
@anon Thank you very much for the help. I tried something like that at first but I was messing things up by trying to take the conjugate of $f + ig$ when it really was the reciprocal that I needed. Would you please add your comment as an answer so that I can accept it and upvote it? – Adrián Barquero Jul 6 '12 at 4:35
up vote 3 down vote accepted

You approach appears to be correct, and it can be finished with the following thought: not only do complex exponentials split into combinations of trigonometric functions, but trig functions also split into combinations of complex exponentials. Indeed:

$$\cos\alpha=\frac{e^{i\alpha}+e^{-i\alpha}}{2},\quad \sin\alpha=\frac{e^{i\alpha}-e^{-i\alpha}}{2i}.$$

This is applicable for not just real $\alpha$, but complex as well. You've deduced $e^{ih(z)}=f(z)+ig(z)$ for some entire function $h$, and taking inverses gives $e^{-ih(z)}=f(z)-ig(z)$, so averaging these two will give you $\cos h(z)=f(z)$ (and similarly, $\sin h(z)=g(z)$).

share|cite|improve this answer
Thank you very much for the help. I'm happy that my argument worked in the end ;-) – Adrián Barquero Jul 6 '12 at 4:40

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