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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)? – user83827 Nov 29 '11 at 14:32
@ccc: the problem statement requires that $f$ be holomorphic. – Jon Nov 29 '11 at 14:49
@ccc: I edited it, sorry for creating the confusion. – VVV Nov 29 '11 at 14:51
Now that it's been edited, it has become a duplicate of this question:… – user83827 Nov 29 '11 at 14:56
Half of it is duplicate. – Graphth Nov 29 '11 at 15:07

To (potentially) cut down on the number of unanswered questions, I'll post GEdgar's answer from the comments as a Community Wiki answer.

Since we're assuming $f$ is holomorphic, then $$f(z)=\sum_{n=0}^\infty a_nz^n$$ holds for all $z\in\Bbb C$, where each $a_n\in\Bbb C$. Since differentiation of a power series is done termwise, then $$f''(z)=\sum_{n=0}^\infty\frac{d^2}{dz^2}[a_nz^n]=\sum_{n=2}n(n-1)a_nz^{n-2}=\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}z^n,$$ so since $f+f''$ is identically zero on $\Bbb C$, then $a_n=-(n+2)(n+1)a_{n+2}$ for all $n\geq 0$.

Let's set $C=a_0$ and $A=a_1$. Since $a_n=-(n+2)(n+1)a_{n+2}$ for all $n\geq 0$, then quick inductive arguments show that for all $k\geq 0$ we have $$a_{2k}=\frac{(-1)^kC}{(2k)!}$$ and $$a_{2k+1}=\frac{(-1)^kA}{(2k+1)!}.$$ Consequently, we have

$\begin{eqnarray*} f(x) & = & \sum_{n=0}^\infty a_nz^n\\ & = & \sum_{k=0}^\infty a_{2k+1}z^{2k+1}+\sum_{k=0}^\infty a_{2k}z^{2k}\\ & = & A\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}z^{2k+1}+C\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}z^{2k}\\ & = & A\sin z+C\cos z. \end{eqnarray*}$

As an addendum, your ansatz $A\sin(bz)+C\cos(bz)$ is not a solution to the differential equation, in general. We may as well assume that $b$ isn't a negative number. Otherwise, we can use odd/even properties of sine and cosine to rewrite it as $A_0\sin(b_0z)+C_0\cos(b_0z)$, where $A_0=-A,C_0=C,b_0=-b$. Note that if $g(z)=A\sin(bz)+C\cos(bz)$ for some $A,C,b\in\Bbb C$, then $$g(z)+g''(z)=(1-b^2)g(z).$$ Consequently, if $g$ isn't identically zero (that is, if $A,C$ not both zero), then $g+g''=0$ if and only if $b=1$ (since $b=1$ is the only nonnegative solution to $1-b^2=0$).

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