complex analysis

**Liav Teichner** · November 27th 2007, 10:26 AM

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)

**Opalg** · November 27th 2007, 12:33 PM

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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