How to prove that f is not a one-to-one map in any neighborhood of $z_0$

Question

Let $U\subset C$ be an open set,$f:U\to C$ be a holomorphic function and $z_0\in U$. Prove that if $f'(z_0)=0$, then f is not a one-to-one in any neighborhood of $z_0$ .

Answer 1

We may assume $f$ is not constant.

Let $\Gamma$ be a simple closed positively-oriented contour contained, together with the region it encloses, in $U$, such that $z_0$ is inside $\Gamma$ but no other zeros of $f(z) - f(z_0)$ or of $f'(z)$ are inside or on $\Gamma$. By the Argument Principle the number of zeros of $f - y$ inside $\Gamma$, counted by order, is $$ N(y) = \dfrac{1}{2\pi i} \oint_\Gamma \dfrac{f'(z)\; dz}{f(z)-y}$$ For $y = f(z_0)$, that number is at least $2$, because the order of $z_0$ as a zero of $f(z) - f(z_0)$ is the least positive integer $m$ such that $f^{(m)}(z_0) \ne 0$. But $N(y)$ is constant as long as $y \notin f(\Gamma)$. If $y \ne z_0$, those zeros are all simple, because at a non-simple zero $f'(z)=0$. So there are at least $2$ distinct $z$ inside $\Gamma$ for which $f(z) = y$, and $f$ is not one-to-one inside $\Gamma$.

Answer 2

We can also argue this way: Suppose WLOG $z_0 = 0, f(0) = 0, f'(0) = 0$ and $f\not \equiv 0$ in some $D(0,r).$ Then $f(z) = z^mg(z)$ for some integer $m>1,$ where $g$ holomorphic in $D(0,r)$ and $g(0) \ne 0.$

By taking $r$ smaller if necessary, we can assume $g$ is never $0$ in $D(0,r).$ Then $g$ has a holomorphic logartithm in $D(0,r),$ which implies $g = h^m$ for some $h$ holomorphic in $D(0,r).$ Thus $f(z) = (zh(z))^m$ in $D(0,r).$

Now $(zh(z))'(0)\ne 0,$ so $zh(z)$ is 1-1 near $0.$ Taking $r$ smaller yet again, we can assume $zh(z)$ is 1-1 in $D(0,r).$ By the open mapping theorem, $zh(z)$ maps $D(0,r)$ onto an open set containing $0.$ Some $D(0,s)$ is contained in this open set. Let $V=(zh(z))^{-1}(D(0,s)\setminus \{0\}).$ Because the function $w\to w^m$ is exactly $m$ to one in $D(0,s)\setminus \{0\},$ $f$ is exactly $m$ to one in $V\setminus\{0\},$ giving the result.