$f, g$ entire functions with $f^2 + g^2 \equiv 1 \implies \exists h $ entire with $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$

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.

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, $\sin h(z)=g(z)$).