calculus - Solving functional equation $f(x)f(y) = f(x+y)$

I'm having some trouble solving the following equation for $f: A \rightarrow B$ where $A \subseteq \mathbb{R}$ and $B \subseteq \mathbb{C}$ such as:

$$f(x)f(y) = f(x+y) \quad \forall x,y \in A$$

The solution is supposed to be the form $f(x) = e^{kx}$ where k can be complex but I don't see how to show it.

Note: I'm trying to solve this equation to find irreductible representation of U(1) but it doesn't really matter I think

Answer

The general solution of the given functional equation depends on conditions imposed to $f$.

We have:

The general continuous non vanishing complex solution of $f(x+y)=f(x)f(y)$ is $\exp(ax+b\bar x)$, where $\bar x$ is the complex conjugate of $x$. Note that this function is not holomorphic.

If we request that $f$ be differentiable then the general solution is $f(x)=\exp (ax)$, as proved in the @doetoe answer.

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Added:

here I give a similar but a bit simpler proof:

Note that $f(x+y)=f(x)f(y) \Rightarrow f(x)=f(x+0)=f(x)f(0) \Rightarrow f(0)=1$ ( if $f$ is not the null function). We have also: $f'(x+y)=f'(x)f(y)$ (where $f'$ is the derivative with respect to $x$).
Now, setting $x=0$ and dividing by $f(y)$ yields:

$$f'(y)=f'(0)f(y) \Rightarrow f(y)=k\exp\left(f'(0)y\right)$$
and, given $f(0)=1$ we must have $k=1$.

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The continuity request can be weakened but, if $f$ is not measurable then, with the the aid of an Hamel basis, we can find more general ''wild'' solutions.

For a proof see: J.Aczél : lectures on functional equations and their applications, pag. 216.