Prove $f'(z)=f(z)$ implies $f(z)=ce^z$ for some $c \in \mathbb{C}$

Question

Suppose that: $$f'(z)=f(z) \text{ for all }z\in\mathbb C.$$ In other words, the complex function $f$ is equal to its own derivative. Prove that there is a constant $c\in\mathbb C$ such that $f(z)=c e^z$; that is $f$ is the exponential function (up to a multiplicative constant).

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So far, I've tried substituting $f'(z)=f(z)$ into the limit definition of $f'(z)$, to no avail. I'm trying to think of what other simple expressions I have relating $f$ and $f'$, but am not having much success.

Answer 1

Let $g(z) = f(z) e^{-z}$. It's the product of two differentiable functions, hence it is differentiable. By the product rule, $g'(z) = f'(z) e^{-z} + f(z) (-e^{-z}) = 0$. Hence $g' = 0$, so $g$ is a constant, say $g(z) = c$. Therefore $f(z) = c e^{z}$.

Note that a priori, you do not know that the complex logarithm of $f$ is well defined. And in fact it's possible that $f = 0$, so in that case it's really not defined at all...

Answer 2

Hint: note that $$ \frac{d}{dz} \ln[f(z)] = \frac{f'(z)}{f(z)} = 1 $$

Answer 3

Re-write as \begin{equation} \frac{df}{dz} = f \end{equation} Thus \begin{equation} \frac{df}{f} = dz \end{equation} from which (and wlog) \begin{equation} \int \frac{df}{f}=\int dz \end{equation} Hence, \begin{equation} \ln f=z+c \end{equation} Hence \begin{eqnarray} f(z)&=&e^{z+c} \\ &=& Ce^{z} \end{eqnarray} Where $C=e^{c}$

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Actually, due to comments below and having another look at the question there is nothing regarding $f$ and how well $f$ is defined. Consequently, one could consider \begin{equation} \frac{d}{dz}\left(e^{-z}f(z)\right) = -e^{-z}f(z)+f'(z)e^{-z} \end{equation} Clearly substituting our condition $f'(z)=f(z)$ we arive at \begin{equation} f(z)(e^{-z}-e^{-z}) = 0 \end{equation} Consequently \begin{equation} e^{-z}f(z)=C \end{equation} where $C$ is a constant. Thus \begin{equation} f(z)=Ce^{z} \end{equation}

Answer 4

At first you should show that the function is holomorphic (by using Cauchy-Riemann differential equations). Then, from a theorem in complex analysis, above differential equation can be solved by the separation of variables. To show the other direction, simply substitute the function $f(z) = ce^z$ into the equation $f'(z) = f(z)$.