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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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uh... I get "page 156 to 176 are not shown in this preview" –  leonbloy Jul 15 '11 at 18:30
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Google Book links are unstable and country-dependent. I cannot even see page 174 from where I am connecting right now. –  Arturo Magidin Jul 15 '11 at 18:32
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Victor: I urge you to improve the quality of your questions. Try to make them self-contained and understandable, please. –  t.b. Jul 15 '11 at 18:35
    
You have also posted it here. –  Ehsan M. Kermani Jul 15 '11 at 18:50
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2 Answers

up vote 2 down vote accepted

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 $|z-z_0|<r$ then $|H(z)-H(z_0)|<\varepsilon$. Plug in $z_0=0$, $\varepsilon=\frac{1}{2}$ and remember that $H(0)=0$.

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