This is from Stein's Complex Analysis page 207 on the chapter about conformal equivalence.
Proof:
We argue by contradiction and supposed that $f'(z_0)=0$ for some
$z_0\in U$. Then$f(z)-f(z_0)=a(z-z_0)^k+G(z)$ 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$f(z)-f(z_0)-w=F(z)+G(z)$, 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, Rouche's 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.
I saw many similar questions on MSE, but I couldn't find a satisfactory answer unfortunately.
I understand we can do the taylor expansion $f$ about $z_0$ and simplify as the above equation. But I don't know why we need sufficiently small $w$. I know Rouche's theorem and that it is used here, but I don't really understand why the above inequality holds and all roots are distinct. I think pictorial explanation will help, and I can't seem to think of any proper one. Can you please explain what's going on in the proof?
Best Answer
The larger $w$ is, the larger the circle needs to be to ensure that $F$ has roots inside of it. If $G$ grows quickly enough, this may push the circle into regions where $|F| \ge |G|$ becomes false. We know exactly how $F$ behaves, but we have very little information about the behavior of $G$.
For simplicity, I'll assume that $z_0 = 0$ and $f(0) = 0$. The general case follows by translation.
There are two things that must be satified. For the $w$ we use, there has to be an $r$ such that
Now $G(z) = z^{k+1}g(z)$ for some holomorphic $g(z)$. Because $g$ is continuous at $z = 0$, it is bounded on a neighborhood. So there is some $M > 0$ and some $R > 0$ such that for all $z \le R, |g(z)| < M$, and therefore $|G(z)| \le |z|^{k+1}M$. To assure both conditions are met, we need to choose $r$ such that $$r <\min\left\{\frac{|a|}M, R\right\}$$ Then choose a $w$ such that $$|w| < \min\left\{|a|r^k, r^k(|a| - rM)\right\}$$
Since $|w| < |a|r^k$, we have $\left|\frac wa\right|^{1/k} < r$, so the roots of $F$ are inside the circle, and since $|w| < r^k|a| - r^{k+1}M$ we have for $|z| = r < R$,
$$|G(z)| \le Mr^{k+1} < |a|r^k - |w| = |a||z^k| - |w| \le |az^k - w| = |F(z)|$$
Since $k \ge 2$, $F$ has at least two roots, which are all within the circle. By Rouche's theorem, so does $f(z) - w = F(z) + G(z)$. Which means there are at least two $z$ for which $f(z) = w$, though this counts multiplicities.
But this works for any $0 < |w| < r$ and as $|w| \to 0$, so do the roots of $f(z) = w$. Since $f'(0) \ne 0$, and $f'$ is continuous, there is some neighborhood of $0$ in which $f' \ne 0$ anywhere. For small enough $w$, the roots $z$ must be in the neighborhood, and since $f'(z) \ne 0$, it cannot be a multiple root. Hence the roots of $f(z) = w$ must be distinct.