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I need to prove that if $f$ is analytic but not one-to-one in the unit disk, then $\exists z,w\in D_1(0)$ such that $|z|=|w|$ and $f(z)=f(w)$.

There is a hint that says to use Argument Principle but I don't know how to use that.

Any help is appreciated.

NECing
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    Naturally, you want to exclude $z=w$. Hint: if $f$ is injective on every circle, then the image of that circle is a Jordan curve. The winding number of a Jordan curve about any point can only be $\pm 1$ or $0$. – 40 votes Jul 23 '13 at 17:31
  • @40votes Thanks. That means $f-f(z_0)$ has only one zero in the unit disk for all $z_0$, which implies $f$ is injective. Correct? – NECing Jul 23 '13 at 17:51
  • Yes. Of course, you should fill in the part about the winding number of Jordan curve. If you can do it and post a solution below, that would be great. – 40 votes Jul 23 '13 at 17:54

1 Answers1

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Following up on comments: suppose the conclusion is false, that is, $f$ is injective on every circle $|z|=r$. Suppose that the value $w\in\mathbb C$ is attained more than once. Since the zeros of $f-w$ are discrete, there is $r$ such that no zeros lie on $|z|=r$ and at least two lie in $|z|<r$.

By the argument principle, the change of argument of $f-w$ on $|z|=r$ must be at least $4\pi$.

However, the winding number of a Jordan curve about any point can be only $0$, $1$ or $-1$. For smooth Jordan curves $\Gamma$ this can be shown as follows: by the Jordan curve theorem, $\mathbb C\setminus \Gamma$ consists of two components; for points in the unbounded component the winding number is $0$. Consider a point $w$ near $\Gamma$. A piece of $\Gamma$ near $w$ looks like a line segment, on which $\arg (f-w)$ changes by about $\pi$. Moving $w$ to the other side of line segment results in $\pi$ being replaced by $-\pi$. Since the total change of argument is a multiple of $2\pi$, it follows that crossing the curve results in $\pm 1$ to the winding number.

With heavier tools (degree theory, Jordan-Schoenflies theorem) one can handle the continuous case too, but only smooth is needed here.

40 votes
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