Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Your notation is quite odd. Normally $z$ is the variable... – Aryabhata Jun 26 '11 at 20:31
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.

share|cite|improve this answer
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

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 $f(0)=1$.

share|cite|improve this answer

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

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

share|cite|improve this answer
FWIW, I find power series to be quite elegant :-) – Aryabhata Jun 26 '11 at 20:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.