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The power series $f(z):= \sum_{\nu \geq 0}a_{\nu}z^{\nu}$ converges in a disc $B$ centered at $0$. For every $z\in B$ such that $2z$ is also in $B$, $f$ satisfies $f(2z) = (f(z))^{2}$. Show that if $f(0)\neq 0$, then $f(z)=\mathrm{exp} bz$, with b:=f'(0)=a_{1}.

I'm completely lost with this problem. Could someone give me a hint?

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    @Chu: because the function is holomorphic, you can differentiate as often as you want. If you have a power series then the $\nu$-th term is the $\nu$-th derivative evaluated at $z=0$. If two holomorphic functions are equal then their power series is equal.2011-03-29

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First you can plug in $z = 0$ to get that $a_0 = 1$. Since $\log(1 + z)$ is analytic on $|z| < 1$, the composition $\log(1 + \sum_{n=1}^{\infty}a_nz^n) = \log(f(z))$ will also be analytic on some disk $D$ centered at $z = 0$. Let $g(z) = \log(f(z))$. Your conditions say that $g(2z) = 2g(z)$ now. Look at the power series of $g(z)$ to show that this equation forces $g(z)$ to be of the form $cz$, and you have enough info to figure out what $c$ is.

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Hint: take the logarithm of your equation and arrive at an equation for $g(z) = \log f(z)$.