Injective holomorphic function is conformal (i.e. nonzero derivative)

Question

STATEMENT: If $f:U\rightarrow V$, where $U,V$ are open subsets of $\mathbb{C}$, is holomorphic and injective, then $f'(z)\neq 0$ for all $z\in U$.

Proof: We argue by contradiction, and suppose that $f'(z_0)=0$ for some $z_0\in U$. Then $$f(z)-f(z_0)=a(z-z_0)^k+G(z)\;\;\;\;\;\;\;\;\text{for all $z$ near $z_0$}$$ with $a\neq 0, k\geq 2$ and $G$ vanishing to order $k+1$ at $z_0$. For sufficiently small $w$ we write $$f(z)-f(z_0)-w=G(z)+F(z)\;\;\;\;\;\;\;\;\text{where $F(z)=a(z-z_0)^k-w$}$$ Since $|G(z)|<|F(z)|$ on a small circle centered at $z_0$, and $F$ has at least two zeros inside that circle, Rouches theorem implies that $f(z)-f(z_0)-w$ has at least two zeros there. Since $f'(z)\neq 0$ for all $z\neq z_0$ but sufficiently close to $z_0$ it follows that the roots of $f(z)-f(z_0)-w$ are distinct, hence $f$ is not injective, a contradiction.

QUESTION: This passage is from Stein's Complex Analysis. My question is how does he conclude that the roots of $f(z)-f(z_0)-w$ are distinct by noting $f(z)\neq 0$ if $z\neq z_0$.

Answer 1

Inside a small circle centered at $z_0$, $f(z)-f(z_0)-w$ has at least two zeros $z_1, z_2$.(Note that if $w\ne 0$, then $z_1\ne z_0, z_2\ne z_0$.) If $z_1=z_2$, then $f(z)-f(z_0)-w=(z-z_1)^2h(z)$.This means $f^\prime (z_1)=0$, which contradicts the fact that $f^\prime (z)≠0$ for all $z≠z_0$ but sufficiently close to $z_0$. Thus the roots of $f(z)−f(z_0)−w$ are distinct.