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

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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.

## marked as duplicate by Hans Lundmark, Najib Idrissi, graydad, Namaste derivatives StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 4 '15 at 15:13

• 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 4 '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 $\frac{d}{dz}\left(e^{-z}f(z)\right)=0$. – Mathemanic Feb 4 '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 4 '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 4 '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 4 '15 at 14:53

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...

• 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 4 '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 4 '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! – Mathemanic Feb 4 '15 at 14:44
• Well then, I did waste a few minutes searching for a duplicate. Try googling "its own derivative site:math.stackexchange.com". – Hans Lundmark Feb 4 '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 4 '15 at 15:07

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}

• How do you know that $\ln f$ is well defined? – Najib Idrissi Feb 4 '15 at 14:31
• @NajibIdrissi You only need it defined locally, I think, which is completely unproblematic. – Potato Feb 4 '15 at 14:35
• Note that $C \neq 0$ therefore, for obvious reasons, you miss the solution $f =0$. – N. S. Feb 4 '15 at 14:41
• Good points well made, I re-edited my answer. – Autolatry Feb 4 '15 at 15:05

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

• How do you know that $\ln f$ is well-defined? – Najib Idrissi Feb 4 '15 at 14:31
• @NajibIdrissi Assume that $f \neq 0$. Certainly, there's an open set on which $f(z) \neq 0$. On this open set, we can select a suitable branch of $\ln$. – Omnomnomnom Feb 4 '15 at 14:34

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