What is the radius of convergence of the power series expansion of $f(z)$ about $z=0$ I was working on the following question from a textbook

Suppose $f(z)$ is analytic at $z=0$ and satisfies $f(z)=z+(f(z)^{2})$. What is the radius of  convergence of the power series expansion of $f(z)$ about $z=0$?

When thinking about this problem, I assumed the naive thing to do would be to be write out the power series expansion as 
$f(z)= f(0)+f'(0)z+ \frac{f^{(2)}(0)}{2!} z^{2} + \cdots$
and then to naively use the relation $f(z)=z+(f(z))^{2}$ to solve for $f^{(k)}(0)$ in terms of $f(0)$.
I looked at the following hint given in the text

Near $0$ the function coincides with one of the branches of $(1 \pm \sqrt{1 - 4z})/2$. The radius of convergence of the power series of either branch is $1/4$, which  is the distance to the singularity at $1/4$.

I am having the trouble connecting the hint to the problem. I am not sure what the text means by "the function coincides with one of the branches of $(1 \pm \sqrt{1 - 4z})/2$". Any help in understanding the hint would be appreciated.
 A: f=z+f² implies f=((1±√(1-4z²))/2)=1/2±(1/2)√(1-4z²). Every branch of this function has a Maclauren series  with radius of convergence =1/2.
A: For each $z \in \Bbb C$ there are at most two possible values $y \in \Bbb C$ satisfying $y = z+y^2$ (hence two possible values for $f(z)$ if it is defined at $z$)  
More precisely, $y = z+y^2 \iff (2y-1)^2 = -4z+1$, so there are $2$ solutions except when $z=1/4$, and then $y=1/2$ is the only solution.
Let $D = \{z \in \Bbb C \mid |z| < 1/4\}$. There is a unique continuous (and in fact holomorphic) function $g : D \to \Bbb C$ such that $g(0) = 0$. And since we know that $f$ is analytic at $0$, $f(0)=0$, and $f$ satisfies the same equation, $f$ has to agree with $g$ on a small neighbourhood of $0$.
In particular, $f$ and $g$ have the same Taylor development $\sum a_kz^k$ at $0$.
But since $g$ is holomorphic on $D$, by Cauchy's differentiation formula we get that $a_k = O(r^{-k})$ for any $r < 1/4$, hence the radius of convergence is at least $1/4$ (if $|B = \sup_{z \in D} |g(z)|$ then we can even deduce that $|a_k| \le B .4^{k}$) .
If the radius of convergence was greater than $1/4$ then it would define an holomorphic function on some open disk containing $1/4$ (and it would still satisfy the equation $f(z) = z + f(z)^2$), but this is impossible because there can't exist a continuous function satisfying that equation defined on any neighbourhood of $1/4$.
Hence the radius of convergence of the power series expansion at $0$ of $f$ is $1/4$.
