$f$ analytic but not one-to-one in unit disk, then $\exists |z|=|w|,f(z)=f(w)$. 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.
 A: 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.
