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This is an exercise from a book I tried:

One would like to find all holomorphic equations that satisfy:$$i) \ f(z)+f''(z) = 0 \text{ in } \mathbb{C} $$$$ii)\ f(z^{2})=f(z)^{2} \text{ in } \mathbb{C}$$

I attempted this:

For i) one can use the Ansatz: $A\sin(bx)+C\cos(bx)$ and this also turns out to be the solution. How to show that these are all equations that satisfy the equation?

For ii) all terms of the form $x,x^2, x^3,\ldots, x^n$ satisfy the functional equation. I don't see any other, so I assume that these are the only ones.

How to show that these are all possible equations that satisfy the functional equations?

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    There are a *lot* of solutions to the second functional equation. Are you leaving out some hypothesis on $f$ (maybe that it's continuous or analytic)?2011-11-29
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    @ccc: the problem statement requires that $f$ be holomorphic.2011-11-29
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    @ccc: I edited it, sorry for creating the confusion.2011-11-29
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    Now that it's been edited, it has become a duplicate of this question: http://math.stackexchange.com/questions/44310/entire-functions-such-that-fz2-fz22011-11-29
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    Half of it is duplicate.2011-11-29
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    @Norbert: rewriting the first equation for real and imaginarty parts gives: $u+iv+ u_{xx}+iv_{xx} = 0 $ So the system of differential equations is: $v = -v_{xx} , u=-u_{xx} $. Is there a method to solve this system?2011-11-29
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    $f + f'' = 0$ ... Assuming holomorphic at zero, deduce the recurrence for the coefficients of the Taylor series, conclude the solution is $A \sin z + C \cos z$ as claimed.2011-11-29
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    @GEdgar Am I right if I think that you mean to look at : $\sum a_{n}(z-z_{0}^{n}$ and $(\sum a_{n}(z-z_{0})^{n})''$ as the taylor series for f and f'', then deducing the reccurence of the coefficients gives: $\sum (z-z_{0})^{n} + (\sum (z-z_{0})^{n})'' = 0 $. How can you conclude from this that the Ansatz solution is then the only one possible?2011-11-29
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    I said "at zero", so use $f(z) = \sum a_n z^n$, differentiate twice, plug into $f+f'' = 0$, find what the coefficients must satisfy. You get a recurrence for the coefficients.2011-11-29

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