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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.

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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

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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.

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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

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