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Thread: complex analysis

  1. #1
    Nov 2007

    complex analysis

    let f be an entire function (holomorphic on the complex plane). every z not on the imaginary axis satisfies: |f(z)|<=(|Rez|^-0.5).
    prove that f is constant.
    (using Liouville's Theorem it's enough to show that f is bounded)
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Suggestion: Cauchy integral formula. Suppose that |z| < r. Then f(z) = \frac1{2\pi i}\oint \frac{f(\zeta)}{\zeta-z}d\zeta, where I want to take the integral round the circle of radius 2r centred at the origin. If we substitute \zeta=2re^{i\theta} then f(z) = \frac1{2\pi i}\int_0^{2\pi} \frac{f(2re^{i\theta})}{\zeta-z}2rie^{i\theta}d\theta. Now estimate the size of this integral, using the facts that |\zeta-z|>r and |f(2re^{i\theta})|\leqslant |\text{re}(2re^{i\theta})|^{-1/2} = |2r\cos\theta|^{-1/2}. That gives

    |f(z)| \leqslant \frac1{2\pi}\int_0^{2\pi} \frac1{\sqrt{2r|\cos\theta|}}\:\frac1r\:2r\,d\thet  a = \frac4{\pi\sqrt{2r}}\int_0^{\pi/2}\frac1{\sqrt{\cos\theta}}d\theta.

    I'm pretty sure that the improper integral of (cos θ)^{-1/2} converges. That would be enough to show that |f(z)|→0 as |z|→∞, and then you can apply Liouville's theorem.
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