Disproving by Picard big theorem I want to solve this question:

Is there a function $f(z)$ analytic on $\mathbb C\setminus \{0\}$ that satisfies $|f(z)|\geq |z|^{-1/2}$ for every $z \neq 0$?

I can solve it with a long way (there isn't exist), but I have a big feeling that Picard big theorem can help me to solve it in a few raws. Can  someone show me how to do it? (if this is true)
I haven't used this theorem ever so I will be happy (if it's not hard) to see a good explanation.
thanks.
 A: Indeed no such function exists.   
a) If $f$ can be extended holomorphically through  $0$, the inequality will fail for small $|z|$ since $|z|^{-1/2}$ tends to $\infty$ when $z$ tends to $0$, whereas $f(z)$ tends to $f(0)$.  
b) If $f$ can be extended meromorphically through  $0$ with a pole of order $k\gt 0$ the function $g(z):=z^kf(z)$ will be entire and satisfy $|g(z) |\geq |z|^{k-1/2} $ so that $h(z)=1/g(z)$  satisfies $|h(z)|\leq |z|^{−k+1/2 }$.
This implies that $h(z)$ can be extended holomorphically at $\infty $ by $h(\infty )=0$ and thus that $g(z)=z^kf(z)=1/h(z)$ has a pole ay $\infty$.
Hence $f(z)$ is meromorphic on the whole extended plane and is thus a rational function , necessarily of the form  $f(z)= \frac {1}{z^k} P(z)$  with $P$ a polynomial  satisfying $P(0)\neq 0 $.
But then:
$\bullet$ if $P$ is not constant our inequality is false at any zero of $P$.
$\bullet$ if $P$ is constant our inequality is false for large $|z|$.
c) If $f(z)$ has an essential singularity at $0$, then in the pointed disk  $D^* $ defined by  $0\lt |z|\lt 1$ the inequality implies  $|f(z)| \gt 1$ .
As you very correctly conjectured, this contradicts the big Picard theorem  according to which $f$ takes all complex values (except maybe one) in that pointed disk,.
