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Starting with just the property $E(x + y) = E(x)E(y)$, one can prove quite a lot of the main properties of the exponential function on real numbers. For example, $E(0) = 1$, and $E'(X) = E(X)$, and $E(nx) = E(x)^n$ all follow from a straight-forward application of definitions.

To move this in to complex analysis and Euler's formula, it seems to me that the key property that needs to be proved is $E(\overline{z})$ = $\overline{E(z)}$.

Is there a nice way to prove this that is in the spirit of this particular line of reasoning?

Edit: For example, Chapter 8 of Baby Rudin follows this reasoning, but resorts to the definition $ E(x) = \Sigma_{n\ge0}(\frac{z^n}{n!})$ to prove the conjugation property.

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closed as off-topic by Alexandre Eremenko, Stefan Waldmann, Neil Hoffman, Robert Bryant, Boris Bukh Dec 24 '18 at 16:04

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    $\begingroup$ You need more than $E(x+y)=E(x)E(y)$ to show that $E'(X) = E(X)$. On one level, you need something to make sure you have the right coefficient. You have $2^{x+y} = 2^x 2^y$, but $(2^x)' \neq 2^x$. On another level, you need to assume that $E(x)$ is differentiable. What assumptions are you allowing here? I'd guess that you are assuming that $E(x)$ is complex-differentiable. $\endgroup$ – user44191 Dec 18 '18 at 0:15
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    $\begingroup$ Doy! Yes, good point. To keep it in this line of reasoning, we could say $E'(x) = \alpha E(x)$ and define $\alpha = 1$ for the first point. Re. the second point, yes, you can assume complex-differentiability. $\endgroup$ – grge Dec 18 '18 at 0:25
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    $\begingroup$ Once you have $E'=E$ and $E(0)=1$ you can appeal to the uniqueness theorem for differential equations (apply it to $E(z)$ and to $\overline{E(\overline{z})}$). $\endgroup$ – Alexandre Eremenko Dec 18 '18 at 0:37
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    $\begingroup$ Note $F(z) = e^{\alpha z}$ satisfies $F(x+y)=F(x)F(y)$, regardless of $\alpha$. In particular, $\alpha$ need not be real, and then $\overline{F(z)} = F(\;\overline{z}\;)$ fails. $\endgroup$ – Gerald Edgar Dec 18 '18 at 1:58