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

In the book(complex variable ;herb silverman), there is a proof about univalent function. My question is how to prove the proposition in special case that $f(z) = f(z_0)$. I take several approachs such as transforming function as $g(z):=f(z)+az$ and applies it similarly but it doesn't work. I really yearn for a method to solve it.

Let $f(z)$ be analytic in a simply connected domain $D$ and on its boundary, the simple closed contour $C$. If $f(z)$ is one-to-one on $C$, then $f(z)$ is one-to-one in $D$.

Proof. Choose a point $z_0$ in $D$ such that $w_0 = f(z_0) ≠ f(z)$ for $z$ on $C$. According to the argument principle, the number of zeros of $f(z)−f(z_0)$ in $D$ is given by $(1/2π)\Delta C \arg{f(z) − f(z_0)}$. By hypothesis, the image of $C $must be a simple closed contour, which we shall denote by $C$. Thus the net change in the argument of $w − w_0 = f(z) − f(z_0)$ as $w = f(z)$ traverses the contour $C$ is either $+2π$ or $−2π$, according to whether the contour is traversed counterclockwise or clockwise. Since $f(z)$ assumes the value $w_0$ at least once in $D$, we must have That is, $f(z)$ assumes the value $f(z_0)$ exactly once in $D$. This proves the theorem for all points $z_0$ in D at which $f(z) ≠ f(z_0)$ when $z$ is on $C$.

If $f(z) = f(z_0)$ at some point on $C$, then the expression $\Delta C \arg {f(z) − f(z_0)}$ is not defined. We leave for the reader the completion of the proof in this special case.

share|cite|improve this question
thanks nameless for improved format. – JY. Dec 9 '12 at 11:38
Sorry, but the proof you give doesn't make sense ... are you sure of all the equal signs? Shouldn't there be some $\neq$ ? – wisefool Dec 9 '12 at 13:40
@wisefool you right. I'am sorry that there are mistyping. – JY. Dec 9 '12 at 15:41
up vote 0 down vote accepted

No matter what the proof should be, but if the problem is the case when, for a given $z_0\in C$ there is $z\in D$ such that $f(z)=f(z_0)$, then the solution might be the following: the image $f(D\cup C)$ is a simply connected domain with boundary $f(C)$, and $f$ is holomorphic on $D$, therefore for any open ball $B$ around $z$, $f(B)$ is an open set around $f(z)$ (open mapping theorem). But $f(z)=f(z_0)\in f(C)$ is on the boundary of $f(D\cup C)$ and this is a contraddiction: $f(z)\in f(C)=bf(D)$ but $f(B)$ is an open set and $f(z)\in f(B)\subseteq f(D)$, therefore $f(z)$ is an inner point of $f(D)$. So, the described situation is impossible.

PS: the fact that $f(C)$ is the boundary of the image is a simple application of the argument principle: any point outside the bounded part of the complement of $f(C)$ has winding number $0$, so it has $0$ preimages.

share|cite|improve this answer
I really appreciate you. I can hardly sleep because of this problem. – JY. Dec 9 '12 at 16:05
but I'm wondering how we could guarantee that the image f(D∪C) is a simply connected domain with boundary f(C). – JY. Dec 9 '12 at 16:55
well, $f(C)$ is a one-to-one image of a closed simple curve, so it is a closed simple curve; $f(D)$ is connected and is contained in $\mathbb{C}\setminus f(C)$. By Jordan theorem (which is essentially a winding number argument), the latter is a disconnected set with two connected components, say $A^+$ and $A^-$. Therefore $f(D\cup C)$ is either $A^+\cup f(C)=\overline{A}^+$ or $A^-\cup f(C)=\overline{A}^-$. One of these two is unbounded, but $D\cup C$ is compact, so its continuous image has to be compact, hence $f(C\cup D)$ is the closure of the bounded component of $\mathbb{C}\setminus f(C)$. – wisefool Dec 11 '12 at 13:45

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.