Let $F$ be analytic and injective on $B(0,1)$. Show that if $B(F(0), |F'(0)|)\subset F(B(0,1))$, then $F(z)= F(0) + F'(0)z$.
I have tried the following: Since $F$ is injective we know that $F'(0)\neq0$ so we can define
$$f(z)=\frac{F(z)-F(0)}{F'(0)}$$
which is still an analytic and injective function.
Then $f(0)=0$ and $f'(0)=1.$ I was able to show that then $B(0,1) \subseteq f(B(0,1))$, using the fact that $B(F(0), |F'(0)|)\subset F(B(0,1))$.
This means, that if we look at the inverse $f^{-1} \colon f(B(0,1)) \longrightarrow B(0,1)$ we can restrict the domain to $B(0,1)$ and look at $g(z)=f^{-1}|_{B(0,1)}$. Since $f$ is analytic and injective, $g$ is also analytic.
I then attempted to apply Schwarz lemma to $g$ but ran into 3 problems:
1) I don't see how I can show that $|g'(0)|=1$ or construct a function from $g$ that satisfies this. Without this I cannot apply Schwarz lemma.
2) Even if I could apply Schwarz lemma I would only get that $g(z)=exp(it)z$ for some real $t$. I would need to show that $t=0$ and I haven't been able to.
3) Even if this all worked out I would get that $F:F^{-1}(B(0,1)) \longrightarrow B(0,1)$ is defined by $F(z)= F(0) + F'(0)z$. But how could I extend this to the whole domain of $F$, that is to the whole unit disc?
So I think I have now worked things out so I am posting a summary as an answer:
1) As pointed out by Christian $g'(0)=1$ since from $f(g(z))=z$ we can deduce $g'(z)=\frac{1}{f'(g(z))}$.
2) If we have $g(z)=exp(it)z$, then $g'(z)=exp(it)$ for all $z$ and from point 1) we get that $t=2\pi k$ for an integer $k$ and so $g(z)=z.$
3) Define $H(z)=F(z)-F(0)-F'(0)z$. Then H is analytic on $B(0,1)$. Since $H(F^{-1}(B(0,1))=0$ we have that there is a non-isolated zero on $B(0,1)$ and by the Identity Theorem $H(Z)=0$ on $B(0,1)$.
The result follows.