3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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:

$$a_n=(n+1)a_{n+1}$$

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 $g'(z)=(f(z)f(-z))'=f'(z)f(-z)-f(z)f'(-z)=0$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.