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I want to prove the following statement:

Define the complex exponential function $\exp$ by $\exp(z)= \sum_{n = 0}^\infty \frac{z^n}{n!}$. Then $\exp$ can be characterized by $$\frac{\mathrm df}{\mathrm dz}=f, f(0)=1, f(z+v)=f(z)f(v) \forall z,w \in \mathbb{C}$$

My idea of the proof: Since the covered topics are the very basic of holomorphic functions (equivalent characterizations of complex differentiability), convergence in $\mathbb{C}$ and power series, we proceed as follows:

We see that $f$ is infinitely many complex differentiable and so we can write it locally around $0$ as a power series $f(x)=\sum_{n=0}^\infty a_n (x-0)^n$. One can differentiate power series termwise to see $a_n=\frac{f^{(n)}(0)}{n!}$, which implies $a_n=\frac{1}{n!}$. We can deduce that $f$ equals $\exp$ locally around $0$. Take any $z \in \mathbb{C}$, then $f(\frac{z}{n})=\exp(\frac{z}{n})$ for $n$ natural number big enough. By applying $f(z+v)=f(z)f(v)$ iteratively, $f(z)=\exp(z)$.

This seems correct, but the problem is we haven't already covered the statement that any holomorphic function can be written locally as a power series. Is there another elementary way of proving this without this statement?

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  • $\begingroup$ It's not clear to me exactly what you mean by saying that $\exp$ "can be defined by" those conditions. If you want to show that those conditions on $f$ imply that $f=\exp$ fine, but that statement is meaningless unless you've already defined $\exp$ by some other means, for example by the power series. Perhaps you mean "can be characterized by"? $\endgroup$ – David C. Ullrich Mar 17 '16 at 13:55
  • $\begingroup$ Two other things you might mean: (i) If $f'=f$ and $f(0)=1$ then $f(u+v)=f(u)+f(v)$, (ii) There is only one function satisfying $f'=f$ and $f(0)=1$. How you prove this depends on exactly what you're trying to prove... $\endgroup$ – David C. Ullrich Mar 17 '16 at 13:58
  • $\begingroup$ @DavidC.Ullrich: I rewrite the statement. Hopefully, it's more clear now. Actually, I do not need to deduce $f(u+v)=f(u)f(v)$ from the other properties and can assume this. Does this proof also works for the complex exponential? $\endgroup$ – bjn Mar 17 '16 at 14:29
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In fact $\exp$ is characterized by just the two conditions $f'=f$, $f(0)=1$. The proof is by a simple technique one learns in an elementary differential equations class: find an integrating factor.

Suppose then that $f'=f$ and $f(0)=1$. Define $$g(z)=\exp(-z)f(z).$$Then the product rule and the chain rule show that $g'=0$. So $g$ is constant: $g(z)=c$. Now $f(0)=1$ implies that $c=1$, so $f(z)=\exp(z)$.

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  • $\begingroup$ It looks like a really simple argument! Can you say how came up with this? (Motivation) $\endgroup$ – bjn Mar 17 '16 at 16:58
  • $\begingroup$ I mentioned that in my post! It's a standard technique for solving first-order linear differential equations - I actually learned it the first time I taught DE. $\endgroup$ – David C. Ullrich Mar 17 '16 at 17:20

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