Complex analysis: Show that there is no analytic function on $B(0,1)$ satisfying $\sqrt{|z|}\leq |f(z)| \leq \sqrt[3]{|z|}$ for all $z\in B(0,1)$

Question

Show that there is no analytic function on $B(0,1)$ satisfying $\sqrt{|z|}\leq |f(z)| \leq \sqrt[3]{|z|}$ for all $z\in B(0,1)$.

The symbol $B(0,1)$ means the open disk with radius $1$ centered at $0$.

I am not sure how to attack those kinds of questions.

Many thanks for any help with this!

Answer 1

$f(0)=0$ so there exists an anlaytic function $g$ in $B(0,1)$ such that $f(z)=zg(z)$. But then $\sqrt {|z|} \leq |z||g(z)$ which implies $|g(z)| \geq \frac 1 {\sqrt {|z|}} \to \infty$ as $z \to 0$, a contradiction.$