Suppose we are given a non-constant function $f: \mathbb{C} \rightarrow \mathbb{C}$ which satisfies the property $f(z+w)=f(z)f(w)$ for all $z, w \in \mathbb{C}$ and is differentiable at the origin, i.e. $f'(0)=a$.
I was having trouble showing such a function must be entire (in fact, it should be a complex exponential function). I've tried proving that the function is holomorphic at an arbitrary point using the limit definition, i.e. I tried showing that we have for $z_{0} \in \mathbb{C}$, the limit below exists :
$$\lim_{z \to z_{0}} \frac{f(z)-f(z_{0})}{z-z_{0}}.$$
This is an old qualifying exam question I've been stuck on, any hints or comments are most welcome.
We have $f(0)=f(0+0)=f(0)^2$ , hence $f(0)=0$ or $f(0)=1$.
Case 1: $f(0)=0$, then f(z)=0 for all $z$ and we are done.
Case 2: $f(0)=1$. For $z \in \mathbb C$ we have
$ \frac{f(z+h)-f(z)}{h}=f(z) \frac{f(h)-1}{h}=f(z) \frac{f(h)-f(0)}{h-0} \to af(z)$ as $h \to 0$.