# the image of finite blaschke product

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.

Hint: Let $$f$$ be a finite Blascke product. If some $$a\in \mathbb D$$ is not in the range of $$f,$$ consider $$\phi_a\circ f.$$
• What is $\phi_a$? – glimpser Apr 26 '19 at 18:28
• 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 – Conrad Apr 26 '19 at 18:34
• $\phi_a(z) = \dfrac{a-z}{1-\bar a z}.$ – zhw. Apr 26 '19 at 18:36