1

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).

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.

Najib Idrissi
  • 54,185
EthanAlvaree
  • 3,412
  • 1
    This has been asked here many times before, but I'm too lazy to search for a duplicate... Anyway, what's $(d/dz)\bigl(e^{-z}f(z)\bigr)$? – Hans Lundmark Feb 04 '15 at 14:31
  • $\frac{d}{dz}\left(e^{-z}f(z)\right)=e^{-z}f'(z)-e^{-z}f(z)$. Substituting $f'(z)=f(z)$, this now says $\frac{d}{dz}\left(e^{-z}f(z)\right)=f(z)\left(e^{-z}-e^{-z}\right)$, so $\frac{d}{dz}\left(e^{-z}f(z)\right)=0$. – EthanAlvaree Feb 04 '15 at 14:34
  • ^--- That is definately the simplest way to go and avoid all possible problems with other approaches like $\log f$, $f'/f$ being well defined etc. – Winther Feb 04 '15 at 14:34
  • I've added words in the question that will make searching for it much easier in the future. Maybe we'll avoid having unfindable duplicates someday. – Najib Idrissi Feb 04 '15 at 14:38
  • (In that question, there is no mentioning of complex $x$, but the proof is exactly the same, so I think it's a duplicate.) – Hans Lundmark Feb 04 '15 at 14:53

4 Answers4

5

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...

Davide Giraudo
  • 172,925
Najib Idrissi
  • 54,185
  • I'm stunned, but I can't find a duplicate. If someone finds one, I'll be happy to vote to close. – Najib Idrissi Feb 04 '15 at 14:35
  • I have seen this question like 4-5 times, but can't find it either now:) Anyway, this is the best solution imo! – Winther Feb 04 '15 at 14:36
  • Very elegant strategy. Would not have thought to use set it up this way and apply the product rule. Thanks for your help! – EthanAlvaree Feb 04 '15 at 14:44
  • 1
    Well then, I did waste a few minutes searching for a duplicate. Try googling "its own derivative site:math.stackexchange.com". – Hans Lundmark Feb 04 '15 at 14:51
  • Ah, thanks. I stubbornly included "complex" in every search, mostly because I was unhappy with all these proofs using complex logarithm... So I didn't find that one. Voted to close. At least this new question will also catch people who search "complex equal own derivative"... – Najib Idrissi Feb 04 '15 at 15:07
2

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 $C=e^{c}$

-----------EDIT-----------------------------------------------------------------

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}

Autolatry
  • 3,016
  • 1
  • 16
  • 16
1

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

Ben Grossmann
  • 225,327
0

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 $f'(z) = f(z)$.

kryomaxim
  • 2,882