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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!

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Your notation is quite odd. Normally $z$ is the variable... –  Aryabhata Jun 26 '11 at 20:31
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3 Answers

up vote 15 down vote accepted

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.

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The nice thing about this proof is that you can use it to prove that $e^z$ is the unique solution of $f'(z)=f(z)$ with $f(0)=1$, since if $f,g$ are solutions, then $h(z) = f(z)g(c-z)$ has the property that $h'(z)=0$, so $h(z)$ is constant, and so $g(c) = h(0) = h(c) = f(c)$. So any solution must be unique. –  Thomas Andrews Jun 26 '11 at 20:53
    
This looks really nice - thank you! One thing I don't understand: "Using the initial condition $e^0=1$ we find that [...] Now let $z=a$" - doesn't this say that $z=a=0$ because $e^ze^{c-z}=e^c$ is only true when $z=0$? –  vonjd Jun 27 '11 at 6:54
    
@vonjd: We proved that $e^ze^{c-z}$ is constant in $z$, so $e^ze^{c-z}=e^c$ for all $z$. –  Corey Jun 27 '11 at 7:07
    
Now I see, very elegant proof - Thank you again! –  vonjd Jun 27 '11 at 7:24
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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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FWIW, I find power series to be quite elegant :-) –  Aryabhata Jun 26 '11 at 20:45
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