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I'd like to show that the image of finite blaschke product from the unit disc on $\mathbb{C}$ onto inself.

I'm sure that this mapping is continuous and the domain, i.e., the unit disc is compact and connected, so is the image under this mapping.

And I know that the unit circle is mapped onto itself and that there is some point in the domain which is mapped to $0$.

But I cannot fill in the gap between 0 and the unit circle.

Could you give me a little hint? I'd like to prove this with my own.

Thank you for your kind answer in advance.

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Hint: Let $f$ be a finite Blascke product. If some $a\in \mathbb D$ is not in the range of $f,$ consider $\phi_a\circ f.$

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  • $\begingroup$ What is $\phi_a$? $\endgroup$ – glimpser Apr 26 '19 at 18:28
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    $\begingroup$ Or use the fundamental theorem of algebra and show that if $B$ is a finite blaschke product of order n and $|\lambda|<1$ or $|\lambda|=1$ the equation $B(z)=\lambda$ becomes a polynomial equation of degree precisely $n$ when clearing denominators and hence it has $n$ roots counted with multiplicity; but by topological reasons, the roots when $\lambda$ is inside the disc are also inside the disc and same on the unit circle; with a little more work one can show that on the unit circle the roots are actually distinct $\endgroup$ – Conrad Apr 26 '19 at 18:34
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    $\begingroup$ $\phi_a(z) = \dfrac{a-z}{1-\bar a z}.$ $\endgroup$ – zhw. Apr 26 '19 at 18:36
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    $\begingroup$ @Conrad Why is the surjectiveness of the mapping related to the distinctness of the roots on unit circle ? $\endgroup$ – glimpser Apr 26 '19 at 19:21
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    $\begingroup$ It is not related directly of course but it is one of the more striking properties of finite blaschke products $\endgroup$ – Conrad Apr 26 '19 at 22:24

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