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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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    @Gortaur: I fixed it; I mean zeros in the upper half plane.2011-09-21

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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!2011-09-21