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Let f be an entire function such that $|f(z)|=1 $ for $|z|=1$. Which are the possible values for $f(0)$? and for $f(17)$ ?

Well, in the case of $f(0)$ we can consider the curve $\gamma(t)= e^{it} $ with domain $[0,2\pi]$. In this case 0 has index 1 in the curve, so by cauchy integral formula: $$ f\left( 0 \right) = \int\limits_{|z| = 1} {\frac{{f\left( z \right)}} {z}dz = \frac{1} {{2\pi i}}\int\limits_0^{2\pi } {\frac{{f\left( {e^{it} } \right)}} {{e^{it} }}ie^{it} dt = } \frac{1} {{2\pi }}\int\limits_0^{2\pi } {f\left( {e^{it} } \right)dt} } $$ I can bound $|f(0)| \le 1$ I don't know if I can get more information about $f(0)$ and for 17 I don't know what can I do, since 17 has index cero in this curve. I can c

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Perhaps the function is constant? – PAD Oct 14 '12 at 6:56
a reasonable guess, but consider like $z^2$ – uncookedfalcon Oct 14 '12 at 7:15
Perhaps $f(z)=z^n$? – PAD Oct 14 '12 at 7:56
So, $f(17)=a 17^n$ for some $n$ where $|a|=1$. – PAD Oct 14 '12 at 8:22

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