# Exercise on proving the uniqueness of an analytic function

Prove that there is a unique analytic function $f: \mathbb{C} \longrightarrow \mathbb{C}$ such that $f'(z) = f(z)$ and $f(0) = 1$.

Hint: Let $g$ be another such function and consider the function $h(z) = f(z)g(-z)$. What do you know about $h(z)$?

My process so far: Let $f(z)=e^z$. I know that $f(z)$ is analytic, and satisfies the above conditions. Taking the hint into consideration I know that $h’(z)=0$ and since $h(z)$ is the product of two analytic functions, its also analytic, implying that $h(z)=k$, and since $f(0)=g(0)=1$ implies that $h(z)=1$. And now I’m stuck. I want to show that this implies $f(z)=g(z)$. Any advise?

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You're almost there but your penultimate sentence is a bit off. So far you proved that $h(z) = f(z)g(-z) = 1$ but $f(z) = e^z$, so $e^z g(-z) = 1$. Can you take it from here? – t.b. Sep 16 '12 at 4:19
I just had a d'oh moment... so g(-z)=1/e^z, implying g(-z)=e^-z, implying g(z)=e^z – Chris Sep 16 '12 at 4:22
Uniqueness seems to follow if you take $h=f/g$, without the need to use the explicit form $e^z$. – timur Sep 16 '12 at 5:05
It suffices to take $h=f-g$, since then $h$ has $0$ of infinite order at $0$. – timur Sep 16 '12 at 5:09

As you wrote, $h(z) = 1$. Hence $e^z g(-z) = 1$. Hence $e^{-z} g(z) = 1$. Hence $g(z) = e^z$ and we are done.
However, if you let $g(z) = e^z$ in the first place, the proof would be a bit easier.