# Analytic function in the punctured plane satisfying $|f(z)| \leq \sqrt{|z|} + \frac{1}{\sqrt{z}}$ is constant

I saw this question on my book (Complex Analysis/Donald & Newman):

Let $f(z)$ be an analytic function in the punctured plane $\{ z \mid z \neq 0 \}$ and assume that $|f(z)| \leq \sqrt{|z|} + \frac{1}{\sqrt{|z|}}$. Show that $f$ is constant.

How I can do it by Cauchy's formula?!

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It might be a standard estimate of $f'(z)$ is zero using Cauchy's formula. –  ougao Dec 17 '12 at 17:15

Since $\lim\limits_{z\to0}zf(z)=0$, Riemann's Theorem says that the singularity at $0$ is removable, so $f$ is entire.

Cauchy's Formula says $$f'(z_0)=\frac1{2\pi i}\oint\frac{f(z)}{(z-z_0)^2}\,\mathrm{d}z$$ where the contour of integration is a circle of radius $R$ about the origin. The integrand is $O\left(R^{-3/2}\right)$ so as $R\to\infty$, the integral vanishes. Therefore, $f'(z_0)=0$ everywhere, so $f$ is constant.

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We shall prove that $f'(z_0)=0$ for all $z_0\in\dot {\mathbb C}$, using nothing but Cauchy's formula.

Fix such a $z_0$ and assume $$0<\epsilon<\min\left\{1,{|z_0|\over2}\right\}\ ,\quad R>1+|z_0|\ .$$ Then $z_0$ lies in the annulus $\Omega:\ \epsilon<|z|<R$, and $f$ is holomorphic in a neighborhood of this annulus. Therefore, according to Cauchy's formula, we have \eqalign{f'(z_0)&={1\over2\pi i}\int_{\partial \Omega}{f(z)\over (z-z_0)^2}\ dz\cr &={1\over2\pi i}\int_{\partial D_R}{f(z)\over (z-z_0)^2}\ dz - {1\over2\pi i}\int_{\partial D_\epsilon}{f(z)\over (z-z_0)^2}\ dz\ .\cr} For $z\in \partial D_R$ the estimates $$|f(z)|\leq2\sqrt{R}\ ,\quad |z-z_0|^2\geq R^2$$ are valid, and for $z\in\partial D_\epsilon$ the estimates $$|f(z|\leq {2\over\sqrt{\epsilon}}\ ,\quad |z-z_0|^2\geq \left({|z_0|\over2}\right)^2\ .$$ It follows that $$|f'(z_0)|\leq {2\over \sqrt{R}} +{4\over|z_0|^2}\ 2\sqrt{\epsilon}\ .$$ As $R$ can be chosen arbitrarily large and $\epsilon>0$ arbitrarily small it follows that $f'(z_0)=0$.

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Nice. Essentially using Cauchy's Formula to prove Riemann's Theorem (+1) –  robjohn Dec 19 '12 at 14:37

Let

$$g(z)=zf(z) \,.$$

Then $g(z)$ is entire.

Indeed $g(z)$ is analytic in the punctured plane, and

$$|g(z)| =|f(z)||z| \leq |z|\sqrt{z}+ \sqrt{z}$$

From here you can easily get that the negative coefficients of the Laurent series of $g(z)$ are $0$. Moreover, $g(0)=0$, which implies that $f(z)$ is also analytic at $z=0$, thus entire.

From there you can simply copy the proof of this problem:

entire function is constant

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