# Analytic functions and open mappings

Can someone explain why complex analytic functions should be open mappings. A complex analytic function $f:D \to \mathbb{C}$ on some open domain $D$ can be thought of as $f:D \to \mathbb{R}^2$, $f(x,y)= (f_1(x,y),f_2(x,y))$. The Cauchy Riemann equations then tell us that the total derivative of $f$ is

$\begin{pmatrix} \partial_x f_1 & -\partial_x f_2 \\ \partial_x f_2 & \partial_x f_1 \end{pmatrix}$

whose determinant is nonzero whenever $f'(z) \neq 0$. Thus the inverse function theorem tells us that $f$ is an open mapping if we knew that $f'(z)$ was never zero.

Can someone explain conceptually why open mapping should hold around a point $z$ where $f'(z)=0$ where the inverse function theorem itself is not enough to show openness?

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Conceptually... without loss of generality, take $f(0) = 0, \; f'(0)=0.$ There is a power series, and since the function is not constant, not all terms are $0.$ There is a first nonvanishing derivative, and the power series begins $$a_n z^n + a_{n+1} z^{n+1} + \cdots,$$ with $a_n \neq 0.$ Near the origin, $$f(z) = a_n z^n \left( 1 + a_{n+1} z + a_{n+2} z^2 + \cdots \right).$$ Now, the quotient $$g(z) = 1 + \frac{a_{n+1}}{a_n} z + \frac{a_{n+2}}{a_n} z^2 + \cdots$$ is nonzero near the origin. So, for some target $b \neq 0$ near the origin, solving $$f(z) = b$$ is the same as solving $$z^n = \frac{b}{a_n g(z)}$$ where the right hand side is nearly constant. The result turns out to be that there are exactly $n$ solutions to $f(z) = b,$ and the ratio of any two of these solutions is very close to an $n$th root of unity.
I could summarize this by saying that, around a zero of order $n,$ the mapping stops being $1$ to $1$ and becomes $n$ to $1.$
Minor typo: the coefficients of $f(z) = a_n z^n + a_{n+1} z^{n+1} + \cdots \neq a_n z^n (1 + a_{n+1} z + \cdots)$. – t.b. Mar 26 '12 at 23:33