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?
 A: 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, 
$$  f(z) = a_n z^n \left(  1 + \frac{a_{n+1}}{a_n} z + \frac{a_{n+2}}{a_n} z^2 + \cdots  \right)=a_n z^ng(z)$$
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. 
This is Theorem 11 on page 131 of Ahlfohrs, section 3.3 on The Local Mapping. 
I could summarize this by saying that,  around a zero of order $n,$ the mapping stops being $1$ to $1$ and becomes $n$ to $1.$
