Thread: 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)

Suggestion: Cauchy integral formula. Suppose that |z| < r. Then $\displaystyle f(z) = \frac1{2\pi i}\oint \frac{f(\zeta)}{\zeta-z}d\zeta$, where I want to take the integral round the circle of radius 2r centred at the origin. If we substitute $\displaystyle \zeta=2re^{i\theta}$ then $\displaystyle f(z) = \frac1{2\pi i}\int_0^{2\pi} \frac{f(2re^{i\theta})}{\zeta-z}2rie^{i\theta}d\theta$. Now estimate the size of this integral, using the facts that $\displaystyle |\zeta-z|>r$ and $\displaystyle |f(2re^{i\theta})|\leqslant |\text{re}(2re^{i\theta})|^{-1/2} = |2r\cos\theta|^{-1/2}$. That gives

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

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.