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?!
Best Answer
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 $\Omega:\ \epsilon<|z|<R$, 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 $$|f(z)|\leq2\sqrt{R}\ ,\quad |z-z_0|\geq R-|z_0|\geq{R\over2}$$ are valid, and for $z\in\partial D_\epsilon$ the estimates $$|f(z|\leq {2\over\sqrt{\epsilon}}\ ,\quad |z-z_0|\geq {|z_0|\over2}\ .$$ It follows that $$|f'(z_0)|\leq {8\over \sqrt{R}} +{8\over|z_0|^2}\ \sqrt{\epsilon}\ .$$ As $R$ can be chosen arbitrarily large and $\epsilon>0$ arbitrarily small it follows that $f'(z_0)=0$.