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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.2012-12-17

3 Answers 3

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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 $$0<\epsilon<\min\left\{1,{|z_0|\over2}\right\}\ ,\quad R\geq\max\{1,2|z_0|\}\ .$$ Then $z_0$ lies in the annulus $\Omega:\ \epsilon<|z|0$ 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)2012-12-19
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    Can't understand why $|z-z_0|^2\ge R^2$. Would someone explain?2015-12-24
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    @Meitar: You are right; I have been sloppy here. The value of $R$ should have been chosen more carefully. Hope it's o.k. now.2015-12-25
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    Are we using $f(z)$ which is undefined at $z = 0$, or the extended entire version of $f(z)$ here?2016-11-09
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    @ChristianBlatter If I used the extended $\overline{f}$, then I only need the contour to be a circle, not annulus, is that correctl?2016-11-09
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    @Misakov: We are using $f$ as given. The origin is not occurring in the argument. Note that the assumption on $f$ is not suggesting that $f$ is extendable to the origin.2016-11-09
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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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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