Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 4 down vote accepted

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 $1.$

share|cite|improve this answer
This is a great answer. +1. – Antonio Vargas Mar 26 '12 at 22:58
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
@t.b., thanks, I have figured out how much of a premium is placed on speed on this site, sometimes things go sideways. – Will Jagy Mar 27 '12 at 0:08
@AntonioVargas, thank you. From your MSE profile, you might enjoy looking at…. I have requested two library books by as I now believe a bounded holomorphic function on the open upper half plane need not extend continuously to the real line, but examples appear to be very difficult. – Will Jagy Mar 27 '12 at 0:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.