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