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Let $f$ be analytic in $B_1(0)$ and suppose that $|f(z)| < M$ for all $z\in B_1(0)$ and $f^{-1}(0) = \{z_1, \ldots, z_k\}$

a) Show that $|f(z)| < M\cdot B(z)$, where $$B(z)=\prod_{i=1}^n\biggl|\frac{z_i-z}{1-\overline{z_i}z}\biggr|\,.$$

b) If $f(0) = Me^{ia}(z_1z_2\ldots z_n)$ is not zero ; find a formula for $f$.

I solved part (a) using maximum principle; however, when I defined new function $g=f/B$, the function is analytic on the unit disc, so why do we have in the question that $f(z_k)=0$ ? I don't think I've used it here.

Can you please help me with part (b)? I don`t know how should I solve it.

Thanks,

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Do you know any LaTeX? If so, I recommend editing this so that it's more clear. As it is, I'm having a hard time parsing it –  Bey Oct 30 '12 at 22:18
    
I`m so sorry, but I really don`t know how to do it. –  Danny Oct 30 '12 at 22:47
    
Here`s the question, but it doesn`t contain part (b) math.stackexchange.com/questions/94122/… –  Danny Oct 30 '12 at 22:58
    
For (a), the whole point of defining the product $B(z)$ is that it has the same zeroes as $f$, whence $g(z)$ is analytic in $D$. (Also, I have submitted some edits to your post that will be peer-reviewed. If/when they go through, please double-check them to make sure I have no significantly altered what you intended to ask.) –  Bey Oct 30 '12 at 23:05
    
Thanks Bey for editing. I really appreciate it and thanks for answering my question. But what about part-b? –  Danny Oct 30 '12 at 23:33
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