Is there an analytic function applying formula? Is there an analytic function $f$ in $\mathbb{C}\backslash \{0\}$ s.t. for every $z\ne0$: $$|f(z)|\ge\frac{1}{\sqrt{|z|}}\, ?$$
 A: If $0$ is an essential singuarity of $f$, then by the Big Picard theorem, $f(z)$ leaves out at most one value in every punctured neighbourhood of $0$.
By assumption, $f(0)\ne 0$ for all $z$, hence we find $z$ with $0<|z|<1$ and $f(z)=1$, so for this $z$ we have  $|f(z)|<\frac1{\sqrt{|z|}}$.
Therefore  $f$ cannot have an essential singularity at $0$, and (as clearly $f\ne0$) it can be written as $f(z)=z^kg(z)$ where $k\in\mathbb Z$ and  $g$ is an entire function with $g(0)\ne0$.
If $k\ge0$, then $f$ is bounded in a neighbourhood of $0$, say $|f(z)|<M$ for $|z|<\epsilon$. If additionally $|z|<\frac1{M^2}$, we obtain $|f(z)|<\frac1{\sqrt{|z|}}$.
If $k<0$, then $g(0)\ne 0$ implies $|g(z)|>a$ for $|z|<\epsilon$ for some $\epsilon>0$, $a>0$. 
By assumption, $|g(z)|\ge|z|^{-k-1/2}$ for $z\ne0$, hence $|g(z)|\ge \epsilon^{-k-1/2}$ for $|z|\ge \epsilon$.
This shows that $z\mapsto \frac1{g(z)}$ is an entire function that is bounded by $\max\{a^{-1},\epsilon^{k+1/2}\} $, hence constant.
Then $f(z)=cz^{k}$ for some $c\ne 0$.
As soon as $|z|>|c|^{-k-1/2}$, we get $|f(z)|<\frac1{\sqrt{|z|}}$.
In all cases, we exhibited some $z$ with $|f(z)|<\frac1{\sqrt{|z|}}$, hence no analytic function $\mathbb C^\times\to\mathbb C$ with $|f(z)|\ge\frac1{\sqrt{|z|}}$ for all $z\ne 0$ can exist.
A: How about this:
Since $f(z)$ is analytic on $\mathbb{C}-\{0\}$, $g(z) = \frac{1}{(f(z))^2}$ is analytic on 
$\mathbb{C}-\{0\}$. Also $\bigg|\frac{g(z)}{z}\bigg| \leq 1$. 
I am sure you will finish the rest (think about the order of the pole at $0$ and use Liouville's Theorem).
