Radius of convergence (complex analysis) Let $f(z)$ be the function $\mathbb{C}\to\mathbb{C}$ so that $f(0)=1$, $f^3-4f+3=z$ for all z and holomorphic in the neighborhood of zero. How to find the convergence radius of the Taylor series at zero? All I know, the radius is equal to the distance to the nearest isolated singularity. This singularity exists because of Picard's theorem: $f'=\frac{1}{3f^2-4}$, and if $f$ is entire function, by Picard's theorem it takes all values or values without 1 point, so at some point $3f^2-4=0$, it is contradiction.
 A: $F(w,z)=w^3-4w+3-z$. Equation will then be written as $F(w,z)=0$. By the inverse function theorem if $F(a,b)=0$ and $\frac{\partial F}{\partial w}(a,b)\neq0$ we can w=w(z) write in the neighbourhood of b, and w(z) is holomorphic in this neighbourhood. For this reason we can have isolated singularity if $\frac{\partial F}{\partial w}(a,b)=3w^2-4=0 \Leftrightarrow w=\frac{2}{\sqrt{3}}, w=-\frac{2}{\sqrt{3}}$ and $z=3+\frac{16}{3\sqrt{3}}, z=3-\frac{16}{3\sqrt{3}}$. We can say exaclty that one of this points is isolated singularity, because if f is entire function, by Picard's theorem it takes all values or values without 1 point, so at some point $3f^2−4=0$, it is contradiction. Now I will prove that $z=3-\frac{16}{3\sqrt{3}}$ is isolated singularity, then it is clear that this is the nearest isolated singularity. Let it not be isolated singularity. This means that $w(3-\frac{16}{3\sqrt{3}})\neq \frac{2}{\sqrt{3}}$. Equation $w^3-4w+3=3-\frac{16}{3\sqrt{3}}$ has 2 roots: $\frac{2}{\sqrt{3}}$ and $\frac{-4}{\sqrt{3}}$, so we have that $w(3-\frac{16}{3\sqrt{3}})=\frac{-4}{\sqrt{3}}$. We know that $w(0)=1$. Define $I:=[3-\frac{16}{3\sqrt{3}},0]$. $w(I)$ is path in complex plane and $w(3-\frac{16}{3\sqrt{3}})=\frac{-4}{\sqrt{3}}\in\{Re<0\}$, $w(0)=1\in \{Re>0\}$, so there is a point x on the segment $I$ so that $Re(w(x))=0$, $w(x)=ib$, b is real. But $w^3(x)-4w(x)+3=x\Leftrightarrow -ib^3-4ib+3=x\Leftrightarrow -ib^3-4ib=x-3=0$, so $x=3$ not in I. It is contradiction.
