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If $f$ is a real function whose derivative is itself, then it is not difficult to see that it is of the form $f(x)=ce^{x}$, $c \in \mathbb{R}$. (If $c=0$, then $f=0$).

Indeed, assume that $g(x)$ satisfies $g'(x)=g(x)$. Take $h(x):=g(x)e^{-x}$. Then by the Chain Rule we have: $h'(x)=g'(x)e^{-x}-g(x)e^{-x}$. Since $g(x)=g'(x)$ we obtain that $h'(x)=0$. Then $h(x)=c$, for some $c \in \mathbb{R}$. Therefore, $c=h(x)=g(x)e^{-x}$, so $g(x)=ce^{x}$.

Does the same proof (with adjustments) hold for complex functions, namely, if $g'(z)=g(z)$, then $g(z)=ce^{iz}$, $c \in \mathbb{C}$?

See this question.

Thank you very much!

user237522
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    The same proof works in $\Bbb C$ for complex differentiable functions. Only the factor $i$ in the exponent is wrong, it should be $g(z) = c e^z$. – Martin R Dec 15 '21 at 15:27
  • @MartinR, thank you for your comment and correction. – user237522 Dec 15 '21 at 15:28
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    another fun way is to notice that complex differentiable are analytic and then to use the power series to get $a_n=(n+1)a_{n+1}, n \ge 0$ so if $a_0=c$ then $a_n=c/n!$ – Conrad Dec 15 '21 at 15:57
  • @Conrad, thank you for your comment. – user237522 Dec 15 '21 at 18:16
  • @Conrad You need to assume that $g$ is analytic on $z = 0$, otherwise need some extra fix. The OP probably wants to say that $g$ is entire but it is not mentioned. – WhatsUp Dec 16 '21 at 19:00
  • @WhatsUp the relation $f'=f$ is linear so wlog we can assume it holds on a small disc centered at zero (otherwise translate); the subtler point is that complex differentiable functions are analytic so the power series trick works – Conrad Dec 16 '21 at 20:32
  • @WhatsUp, thank you. I do not mind if $g$ is not entire. $g$ could be holomorphic on $\mathbb{C}$ minus a finite set (like quotient of two polynomials). – user237522 Dec 16 '21 at 21:42

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Indeed, Martin R and Conrad in the comments are correct. The proof works exactly the same for complex functions, since the product rule is the same, and the exponential function has the same differentiability properties in the complex numbers. And as Conrad pointed out, this is also easy to see with the Maclaurin expansion.

Angel
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