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Suppose $B$ is a Blaschke product with at least one zero off the origin, and $B(z)=\sum_{k=0}^\infty {c_kz^k}$. Is it possible that $c_k\ge0$ for all $k=0,1,\ldots$?

My try: Since $B(z)$ takes real values on the real axis, by Schwarz reflection principle, we know $B(\bar z)=\overline{B(z)}$. This happens if and only if, for every zero of $B$, its conjugate must be a zero of $B$ too.

I have no idea how to use the assumption $c_k\ge0$. Can anyone give a hint? Thanks.

By the way, I guess the answer is negative, isn't it?

Note: A Blaschke product is referred to the infinite/finite product $$B(z)=z^k\prod_n\frac{z-\alpha_n}{1-\bar\alpha_nz}\frac{|\alpha_n|}{\alpha_n},$$ where $k\in\mathbb N$ and $\sum_n(1-|\alpha_n|)<+\infty$, $\color{red}{0<|\alpha_n|<1}$.

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  • $\begingroup$ You say $c_k\ge0$. Is it assumed that $c_k$ is real? $\endgroup$ – user141614 Apr 7 '15 at 9:17
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Try $$\alpha_1 = 4,\ \alpha_2 = 3, \alpha_3 = 2 i,\ \alpha_4 = -2 i$$ $$ \eqalign{ B &= {\frac { \left( z-4 \right) \left( z-3 \right) \left( z-2\,i \right) \left( z+2\,i \right) }{ \left( 1-3\,z \right) \left( 1-4\, z \right) \left( 1+2\,iz \right) \left( 1-2\,iz \right) }}\cr &= 48+308\,z+1404\,{z}^{2}+6237\,{z}^{3}+27516\,{z}^{4}+117348\,{z}^{5}+ 488424\,{z}^{6}+ \ldots} $$

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  • $\begingroup$ Sorry, I forgot to write a restriction for $\alpha_n$. Now I have corrected it. But your counterexample is nice. $\endgroup$ – Eclipse Sun Apr 7 '15 at 15:39

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