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?
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 $1-b^2=0$).