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!
On
Let $g(z) = e^{a+z}/e^a$. Then $g'(z) = g(z)$ and $g(0) = 1$. So $g(z) = e^z$, and we have that $e^{a+z} = e^ae^z$.
That assumes that $e^z$ is the only solution to $f'(z)=f(z)$ with $f(0)=1$.
On
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
$$f(a+b) = f(a)f(b)$$
If you define $e^z$ as the unique solution to the ODE $f'(z)=f(z)$ with initial condition $f(0)=1$, then you have by the product rule: $$ (e^ze^{c-z})'=e^ze^{c-z} + e^z(-e^{c-z})=0.$$ Thus $e^ze^{c-z}$ is a constant. Using the initial condition $e^0=1$ we find that $e^ze^{c-z}=e^c$. Now let $z=a$ and $c=a+b$ and the result follows.