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please see here(p.174-175 Elementary real and complex analysis By Georgiĭ Evgenʹevich Shilov):

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Question is, why is $\displaystyle |H(z)| \lt 1/2$ true?

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    You have also posted it [here](http://www.artofproblemsolving.com/Forum/viewtopic.php?t=418421&p2360294#p2360294).2011-07-15

2 Answers 2

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It is the definition of continuity in $0$. For any $\epsilon$ you can find $\delta$ such as $z$ lies in the disc centered in $0$ with radius $\delta$ implies $H(z)$ is in the disc centered on $H(0)=0$ with radius $\epsilon=1/2$.

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It follows from continuity of polynomials: If $H$ is continuus at $z_0$ then for any $\varepsilon>0$ there exists $r>0$ such that if $|z-z_0| then $|H(z)-H(z_0)|<\varepsilon$. Plug in $z_0=0$, $\varepsilon=\frac{1}{2}$ and remember that $H(0)=0$.