Entire extension of $f(x+y)=g(x)g(y)-h(x)h(y)$ I am currently working on the following practice question for complex analysis;
Assume $f(x+y)=g(x)g(y)-h(x)h(y)$ for all $x,y\in \mathbb{R}$ and some entire functions $g,h$. Show that there exists a unique entire function $F$ such that
$F(z+w)= g(z)g(w)-h(z)h(w)$ for all $z,w \in \mathbb{C}$.
As usual, uniqueness is easy to show. However, I am struggling with the existence part. So far I am thinking of defining $F$ as follows;
$$
F(z)=g(z)g(0)-h(z)h(0)   \ \ \ z\in\mathbb{C}.
$$
This function is certainly entire, but I am not sure that it satisfies the property
$F(z+w)= g(z)g(w)-h(z)h(w)$ for all $z,w \in \mathbb{C}$.
Another definition of $F$ which I came up with is
$$
F(z)= g(z/2)^2-h(z/2)^2
$$
but again i'm not sure how to impose the required properties.
Any hints are appreciated. I have added my proof as an answer below.
 A: This is the proof I ended up coming up with for this question.
Assume we have that
$$
f(x+y)=g(x)g(y)-h(x)h(y)
$$
for all $x,y\in \mathbb{R}$, for some entire functions $g,h$. We will show that there exists an entire function $\tilde{f}$ such that
$$
\tilde{f}(z+w)=g(w)g(w)-h(z)h(w)
$$
for all $z,w\in \mathbb{C}$.
To see this we define the function $\tilde{f}$ by;
$$
\tilde{f}(z)= g(z)g(0)-h(z)h(0) \ \ \ z\in \mathbb{C}.
$$
Then by the algebra of differentiation $\tilde{f}$ is also differentiable on all of $\mathbb{C}$, hence $\tilde{f}$ is entire. Notice
$$
f(x) = g(x)g(0)-h(x)h(0)
$$
for all real $x$. So we have that $f(x)= \tilde{f}(x)$ for all $x\in \mathbb{R}$. Therefore, we can conclude
$$
\tilde{f}(x+y)=g(x)g(y)-h(x)h(y)
$$
for all $x,y\in \mathbb{R}$. Now we fix an arbitrary $y \in \mathbb{R}$ and consider $\tilde{f}(z+y)$ for $z\in \mathbb{C}$ (here we view $z$ as a variable). If $z \in \mathbb{R}$ we have
$$
\tilde{f}(z+y)=f(z+y) = g(z)g(y)-h(z)h(y)
$$
and so by the uniqueness theorem we can conclude that 
$$
\tilde{f}(z+y)= g(z)g(y)-h(z)h(y)
$$ 
for all $z \in \mathbb{C}$ and since $y$ was arbitrary this also holds for all $y\in \mathbb{R}$. Now fix an arbitrary $z\in \mathbb{C}$ and consider $w \in \mathbb{C}$( here we view $w$ as a variable). Then if $w \in \mathbb{R}$ we have
$$
\tilde{f}(z+w)= g(z)g(w)-h(z)h(w).
$$
By the uniqueness theorem  we can then conclude that this holds for all $w \in \mathbb{C}$. Since $z$ was arbitrary we can also conclude that
$$
\tilde{f}(z+w)= g(z)g(w)-h(z)h(w)
$$
for all $z,w\in \mathbb{C}$. This however only gives existence. To see uniqueness suppose we have two entire functions $\tilde{f}$ and $\hat{f}$ such that
$$
\tilde{f}(z+w)= g(z)g(w)-h(z)h(w)\ \ \ \text{ and }\ \ \ \hat{f}(z+w)= g(z)g(w)-h(z)h(w)
$$
for all $z,w\in \mathbb{C}$. Then in particular if we restrict $z \in \mathbb{R}$ and $w =0$ we find that $\tilde{f}$ and $\hat{f}$ agree on the real line and thus by the uniqueness theorem $\tilde{f}$ and $\hat{f}$ are the same functions.
