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!

## Best Answer

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.