Power series and $f(z) = z + f(z^2)$ 
Use power series $f(z) = \sum_{j=0}^\infty a_j z^j$ to solve the
  functional equation $f(z) = z + f(z^2)$.

First, I'm not really sure what the question is asking for. Usually, when I see a functional equation, it's something like $f(\frac{x+7}{2})$ and I have to solve for $f(x)$. In this case, I already have $f(z)$. It the problem asking to get rid of $f(z^2)$?
I can see that $f(z) = z + z^2 + z^4 + \cdots$.
Also that $f(z) = z+ \sum_{j=0}^\infty a_j z^{2^j}$.
So it seems that $f(z) = z + \sum_{j=1}^\infty a_j z^{2^j}$.
Thus, I have gotten rid of $f(z^2)$, which I thought was the point.
However, the answer is the following:

$f(z) = a_0 +\sum_{j=0}^\infty z^{2^j}$, $|z| < 1$

How can I make that leap from $z$ to $a_0$ and make the power series index from 0 instead of 1? How was I supposed to determine the bound for $|z|$?
 A: To solve a functional equation means to find $f$.
You do not already have $f$ because you don't kow the coefficients $a_j$.
The coefficient of $z^j$ on the left is clearly $a_j$, the coefficient of $z^j$ on the right is $=1$ if $j=1$, $=a_{j/2}$ if $j$ is even, $=0$ if $j$ is odd $\ge3$.
It seems you already managed to determine all $a_j$ from this, namely $a_j=\begin{cases}1&\text{if $j=2^k$ fore some $k\ge 0$}\\0&\text{otherwise}\end{cases}$. However, looking closely at the conditions imposed on the $a_j$, we notice that $a_0$ can still be arbitrary. Hence the solution
$$f(z)=a_0+\sum_{k=0}^\infty z^{2^k}.$$
To complete the answer, it is required to check where this series converges. The radius of convergence is clearly $1$ (by Cauchy-Hadamard) and there is no convergence for $z=\pm1$, hence the bound $|z|<1$. Indeed, $f$ cannot bedefined at $z=1$ as the functional equation reads $f(1)=1+f(1)$, which is absurd. Likewise, $f(-1)=-1+f(1)$ makes $f(-1)$ undefined.
A: The kicker is that $$\sum_{k=0}^\infty a_kz^k=f(z)=z+f(z^2)=z+\sum_{j=0}^\infty a_jz^{2j}=a_0+z+\sum_{k\ge 2,k\text{ even}}a_{k/2}z^k.$$ Equating coefficients, we find that $a_k=0$ for all odd $k\ge 3,$ and we find that for even $k\ge 2,$ $a_k=a_{k/2}.$ Putting these facts together, we see that $a_k=0$ when $k>0$ has any prime factor other than $2.$ Then, noting that $a_1=1$ (why?), we have that $a_{2^j}=1$ for all $j\ge 0,$ whence $$f(z)=a_0+\sum_{j=0}^\infty z^{2^j}.$$ This is rather close to what you came up with, but you forgot the constant term (about which we can't draw any conclusions)!
To find the radius of convergence, try applying the root test.
