TITLE: Proof for law of complex exponents using only differential equation
QUESTION [8 upvotes]: I just read that an elegant proof exists that the law of exponents also holds for complex numbers ($a,b,z$ all complex): $$e^{a+b}=e^ae^b,$$ which only uses the definition that $$y=e^{zt}$$ is a solution to $$dy/dt=zy,$$ with initial condition $y(0)=1$, so in particular $e^z=y(1).$
I can only find a proofs which use the trig-representation of complex numbers.
Can anybody help?
Thank you!

REPLY [5 votes]: Another way is to see that any $f: \mathbb{C} \to \mathbb{C}$ satisfying $f'(z) = f(z)$ and $f(0) = 1$ is analytic in $\mathbb{C}$ (entire) and admits a power series representation
$$ f(z) = \sum_{n=0}^{\infty} a_n z^n$$
The fact that $f'(z) = f(z)$ and $f(0) = 1$ easily give us
$$f(z) = \sum_{n=0}^{\infty} \frac{z^n}{n!}$$
Now it is easy to verify that $f$ indeed satisfies the above differential equation and initial conditions (and hence is the unique function) and that
$$f(a+b) = f(a)f(b)$$