If $f:U\to V$ is holomorphic and $f'(z)\neq 0$ for all $z\in U$, then$f$ is locally bijective.

Question

I am trying to solve the following problem:

Let $f:U\to V$ be a holomorphic function such that $f'(z)\neq 0$ for all $z\in U$. Show that for all $z_0\in U$, there exists a disc $D_\varepsilon(z_0)\subseteq U$ such that $f:D_\varepsilon(z_0)\to f(D_\varepsilon(z_0))$ is bijective.

Atempt:

I am trying to use Rouche's Theorem, but I am stuck on a particular step. Let $z_0\in U$. Since $f'(z_0)\neq 0$, there is a disc $D_r(z_0)\subseteq U$ such that $$f(z)=f(z_0)+(z-z_0)h(z),\quad\forall z\in D_r(z_0)$$ where $h$ is holomorphic on $D_r(z_0)$ and $h(z)\neq 0$ for all $z\in D_r(z_0)$.

Let $0<\varepsilon<r$ to be defined later and fix $w\in D_\varepsilon(z_0)$. To show that $f$ is injective in $D_\varepsilon(z_0)$ is to show that the function $f(z)-f(w)$ has exactly one zero in $D_\varepsilon(z_0)$. But $$F(z):=(z-z_0)h(z)$$ has exactly one zero in $D_\varepsilon(z_0)$, so we might want to apply Rouche's Theorem with $F$ and $$G(z):=f(z)-f(w)-(z-z_0)h(z)=f(z_0)-f(w)=-(w-z_0)h(w)$$ to conclude that $F$ and $F+G$ have the same number of zeros and hence $f$ is injective. But that means that we need to find $0<\varepsilon<r$ such that $|G(z)|<|F(z)|$ on $\partial D_\varepsilon(z_0)$. That is, $$|w-z_0||h(w)|<|z-z_0||h(z)|$$ for all $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$.

Can we find such $\varepsilon>0$?

It seems intuitive since if we expand $h$ in a power series at $z_0$, $h(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$, then $$|w-z_0||h(w)| = |w-z_0|\sum_{n=0}^\infty a_n|w-z_0|^n < |z-z_0|\sum_{n=0}^\infty a_n|z-z_0|^n.$$ for $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$. But this is of course not a proof.

Answer 1

I am trying to use Rouche's Theorem ...

Here is a possible approach:

If $f(z_0) = w_0$ and $f'(z_0) \ne 0$ then for sufficiently small $r > 0$:

• $f(z) \ne w_0$ for $0 < |z-z_0| \le r$.

Then $m = \min \{ |f(z) - w_0| : |z - z_0| = r \}$ is positive, and for $|w-w_0| < m$ and $|z - z_0| = r$ $$ |(f(z) - w) - (f(z) - w_0)| = |w- w_0| < m \le |f(z) - w_0| \, . $$ Now Rouché's theorem implies that $f(z) - w$ and $f(z) - w_0$ have the same number of zeros in $|z-z_0| < r$.

It follows that $f$ assumes every value $w$ with $|w-w_0| < m$ exactly once in $|z-z_0| < r$, and therefore is injective in a neighborhood of $z_0$.

Answer 2

In fact this question does not need Rouché's Theorem because we know the nonzero derivative.

We view $\mathbb C$ as $\mathbb R^2$. Then the Jacobian matrix of $f$ at $z$ is: $$\begin{bmatrix}a & -b\\b & a\end{bmatrix},$$ where $f'(z)=a+bi$.

Note that this is the matrix of multiplication by $a+bi$ under standard basis.

The determinant is just the norm square of $f'(z)$, which is nonzero by assumption. Hence the inverse function theorem can be applied.

One remark is that Rouché is extremely useful when we do not know the derivative is nonzero. For example, in the proof of open mapping theorem and the fact that injective holomorphic function has nonzero derivative.

Answer 3

You can use tke inverse fuction theorem, look here: http://en.wikipedia.org/wiki/Inverse_function_theorem