I am reading E.Stein's Complex Analysis. I am stuck at some detail of his proof of holomorphic injective function has non-zero derivative.(I know this statement has some proof in the website, but I am not seeking some proof, but understanding of Stein's Proof, so please do not closed my question)
Here is the notation: $F:U\to V$ is holomorphic and injective, $U,V$ are open.
I got the idea of his proof. Suppose $f'(z_0)=0$ for some $z_0$ in $U$. By Taylor expansion, we have$$f(z)-f(z_0)=a_k(z-z_0)^k+\sum_{n=k+1}^{\infty}a_n(z-z_0)^n$$where $2\le k$ and $a_k\neq0$ near $z_0$, say $B_r(z_0)$. Choose $w+f(z_0)=f(z_1)$ for some $z_1 \in B_r(z_0)$. The idea is to prove $w$ has $2$ pre-images in $B_r(z_0)$. i.e $f(z)-f(z_0)-w=F(z)+G(z)$ has $2$ roots, where $F(z)=a_k(z-z_0)^k-w, G(z)=\sum_{n=k+1}^{\infty}a_n(z-z_0)^n$. Since $G$ has degree larger than $F$, then there exists $r'<r$ such that $|F(z)|>|G(z)|$ on $C_{r'}$, also since $2\le k$, $F$ has at least $2$ zeros on $\mathbb C$ By Rouché's theorem, we have $F+G$ has at least $2$ zeros inside of $C_{r'}$. Since $f'$ has isolated zero at $z_0$, there exists $0<r''<r'$ such that $F'+G'\neq0$ on $B_{r''}(z_0)$, so they cannot have repeated zero on $B_{r''}$.
Here is my question. How can we make sure, the zeros are inside of $B_{r''}\subseteq B_{r'}$? If they are not in that. we cannot claim $F+G$ has $2$ zeros and distinct zeros in $B_{r''}$.
Any Help Will Be Appreciated.