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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
up vote 1 down vote accepted

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.

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We're assuming g(z) is not e^z, thus the contradiction. – Chris Sep 17 '12 at 7:20
@Chris Who are "we"? – Makoto Kato Sep 17 '12 at 7:24
It's a phrase mathematicians use because they're lonely. But more precisely, the proof is asking for uniqueness, i.e. wts f(z)=/g(z) results in a contradiction or equivalently that the existence of any g(z) implies g(z)=f(z) – Chris Sep 17 '12 at 7:43
@Chris We don't need to prove it by showing a contradiction as I proved it without a contradiction. – Makoto Kato Sep 17 '12 at 7:46
The second part of your answer asks to assume g(z)=e^z, when f(z)=e^z is already an assumption. Take that part out, and your answer will be correct. – Chris Sep 17 '12 at 7:50

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