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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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    It suffices to take $h=f-g$, since then $h$ has $0$ of infinite order at $0$. – 2012-09-16

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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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    Ahh ok thank you. – 2012-09-17