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.