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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!

  • 0
    Your notation is quite odd. Normally $z$ is the variable...2011-06-26

3 Answers 3

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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.

  • 0
    Now I see, very elegant proof - Thank you again!2011-06-27
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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 $f(0)=1$.

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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

$f(a+b) = f(a)f(b)$

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    $F$WIW, I find power series to be quite elegant :-)2019-05-21