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If $f$ is an entire function with $|f(z)|>|f(\bar{z})|$ for all complex numbers $z$ in the upper half plane. How does this imply that $f$ has no zeros in the upper half plane?

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Where? If you're talking about zeros in the upper half plane then $f(z) = 0$ implies $0>|f(\bar z)|$ which cannot hold. –  Ilya Sep 21 '11 at 16:22
    
@Gortaur: I fixed it; I mean zeros in the upper half plane. –  lenov Sep 21 '11 at 16:46

1 Answer 1

The answer is simple: for each $z$ in the upper half plain $|f(z)|>|f(\bar z)|\geq 0$ hence $f(z)\neq 0$.

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WOW, just that! –  lenov Sep 21 '11 at 19:53

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