Existence of Analytic and non constant function

Question

Can we construct a non constant analytic function $f(z)$ on open unit disc such that:

$\frac{1}{\sqrt{n}} \lt |f(\frac{1}{n})| \lt \frac{2}{\sqrt{n}} \forall n \in \mathbb N $

My approach: I think no. As $n$ approaches infinity, we observe that $f(0)=0$. So $f$ must have a zero of some order at $z=0$, which means $f(z)=z^m g(z)$ where $m$ is the order of the zero and g(z) is analytic in some nbd of $z=0$ and $g(0)\neq 0$. Since $g(z)$ is continuous at zero so it must be non zero im some nbd of zero, but what to do next to arrive at a possible contradiction?

I tried substituting $g(z)$ in the given inequality but that did not help. Please suggest. Thanks in advance.

Answer 1

If $f(z)=z^mg(z)$, then $f\left(\frac1n\right)=\frac1{n^m}g\left(\frac1n\right)$ and therefore $\lim_{n\to\infty}n^m\left\lvert f\left(\frac1n\right)\right\rvert=\lvert g(0)\rvert$. But$$n^{m-1/2}=n^m\frac1{\sqrt n}<n^m\left\lvert f\left(\frac1n\right)\right\rvert$$and therefore, since $m\geqslant1$,$$\lim_{n\to\infty}n^m\left|f\left(\frac1n\right)\right|=\infty.$$But it was already established that $\lim_{n\to\infty}n^m\left|f\left(\frac1n\right)\right|=|g(0)|$.