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It is well known that a finite Blaschke product $$B(z)=\prod_{k=1}^n\left(\dfrac{z-a_k}{1-\overline{a_k}z}\right),\,\,\,\,\,\,\,\forall a_k\in\Bbb{D}$$ has exactly $n$ distinct per-images of any $\lambda\in\partial\mathbb{D}.$
Let us take them $(x_k)_{k=1}^n\in\mathbb{D},$ where $\arg x_1\lt\arg x_2\lt\cdots\lt\arg x_n\lt2\pi+\arg x_1.$

How can I prove that each arc $[x_k, x_{k+1})$ mapped bijectively on to the unit circle?

I thought we can use to fact that the derivative of $B$ never vanishes on the unit circle. $$|B^{'}(z)|=\sum_{k=1}^n\dfrac{1-|a_k|^2}{|z-a_k|^2}\gt 0,\,\,\,\,\,\,\,\forall z\in\partial\Bbb{D}.$$ But I could not figure out a relationship between these properties.
Can anyone help me to prove this?
At least a hint regarding this?

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    $\begingroup$ Since $B'$ doesn't vanish on the unit circle, the restriction of $B$ to it is a local homeomorphism (local diffeomorphism, but we don't need that). By compactness, it's a covering. $\endgroup$ – Daniel Fischer Oct 11 '16 at 18:43
  • $\begingroup$ Thank you for your reply. But this is little bit advanced for me. Could you please explain it bit more? $\endgroup$ – Bumblebee Oct 11 '16 at 18:48
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    $\begingroup$ Okay, let's forget about coverings. The map $t \mapsto B(e^{it})$ always traverses the unit circle in the same direction (positive orientation), for if it would change direction somewhere, say at $t_0$, that would force $B'(e^{it_0}) = 0$. $\endgroup$ – Daniel Fischer Oct 11 '16 at 18:52
  • $\begingroup$ I think , It makes some sense. Since $[x_k, x_{k+1})$ is connected $B([x_k, x_{k+1}))$ should be connected. Therefore it should be either an arc or the unit circle. If it is an are, since $B(x_k)=B(x_{k+1}),$ somewhere in $[x_k, x_{k+1})$ $B$ should change its direction. But still I am not clear why $B'(e^{it_0}=0$ implies $B$ changes its direction at $t_0$ $\endgroup$ – Bumblebee Oct 11 '16 at 19:10
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    $\begingroup$ It's the other way round, a change of direction would imply a zero of $B'$. Since there are no zeros of $B'$ on the unit circle, there's no change of direction. $\endgroup$ – Daniel Fischer Oct 11 '16 at 19:44
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The abstract argument by Daniel is more elegant. If you want to do it by hand, however, here is a suggestion: Without taking absolute values we have:

$$ \frac{B'(z)}{B(z)} = \sum_{k=1}^n \left( \frac{1}{z-a_k} + \frac{\overline{a}_k}{1-\overline{a}_k z} \right) = \sum_{k=1}^n \frac{1-|a_k|^2}{(z-a_k)(1-\overline{a}_k z)} $$ The path $z=e^{it}$ then maps to
$$ \frac{B'(z)}{B(z)} = e^{-it} \sum_{k=1}^n \frac{1-|a_k|^2}{|e^{it}-a_k|^2} $$ or if we pose $B(t)=e^{i\phi(t)}$: $$\phi'(t) = -i \frac{1}{B}\frac{dB}{dt}=z \frac{B'(z)}{B(z)} = \sum_{k=1}^n \frac{1-|a_k|^2}{|e^{it}-a_k|^2} >0 $$ The argument of $B(z(t))$ is thus monotonously increasing with $t$. The fact that it winds around $n$ times may be seen from calculating $$ \int_0^{2\pi} \phi'(t)\frac{dt}{2\pi} = \int_0^{2\pi} \frac{1}{B}\frac{dB}{dt} \frac{dt}{2\pi i}= \oint \frac{B'(z)}{B(z)}\frac{dz}{2\pi i} = \oint \sum_{k=1}^n \left( \frac{1}{z-a_k} + \frac{\overline{a}_k}{1-\overline{a}_k z} \right) \frac{dz}{2\pi i} = n $$ using the first formula and residue calculus (there are $n$ poles inside the contour).

A simpler argument is that it must be an integer multiple of $2\pi$ and depend continuously upon the $a_k$'s. So now let all $a_k\rightarrow 0$ and you end up with $B(z)=z^n$ for which the result is obvious.

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  • $\begingroup$ Thank you for your answer. I should edit my problem according to you comment :) $\endgroup$ – Bumblebee Oct 11 '16 at 19:53

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