I have a Complex Analysis assessment question about holomorphic functions:

Let f be a function on a plane and satisfies $f'(z) = f(z)$ and $f(0) = 1$

i) Give an example of a function with this property.

ii) By considering $g(z) = f(z)f(-z)$, show that $f(-z) = 1/f(z)$ for all z. (You may assume that a function which is holomorphic on the plane with zero derivative is constant).

I have an answer for part i): $e^z$. (in fact, I can't actually think of any other functions that satisfy these conditions, do any exist?)

Part ii), however, I have no idea how to answer. I can obviously show it using $f(z) = e^z$ and substituting it into the equation, but I'm not sure that's the right way of going about it. Could anyone offer some help?

Thanks a lot.


Hint: Try to calculate $g'$, using $f=f'$ you should reach a simple result that you can use the allowed assumption on.


Let $f(x)=\sum_{n=0}^{\infty}a_nx^n$, then $f'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$, so you have:


By $f(0)=1$ you have $a_0=1$ Using this reccurence relation and condition $a_0=1$ you have $f(x)=e^{x}$.

In second part use Leibniz formula $g'(z)=(f(z)f(-z))'=f'(z)f(-z)-f(z)f'(-z)=0$.

  • $\begingroup$ Hey agha and @Henrik, thanks for both of your replies. I've done what you both suggested, and found $g'(z) = 0$. Which means that $g$ is constant, using the assumption we are given, right? So let's say $g$ is some constant $c$. So we can say that $g(z) = c = f(z)f(-z)$, as was given in the question. But how do we know that this constant, $c$, equals $1$? To show that $f(-z) = 1 / f(z)$, surely $g(z)$ must equal $1$. Forgive my stupidity here. $\endgroup$ – fhodngodfn Jan 30 '15 at 15:08
  • $\begingroup$ Using that $f(0)=1$ you have $g(0)=c=f(0)f(0)=1$. $\endgroup$ – Henrik Jan 30 '15 at 15:59
  • $\begingroup$ Ah, of course. Thanks so much to you and @agha. Have a great day to you both! $\endgroup$ – fhodngodfn Jan 30 '15 at 16:17

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