Showing that a family is not normal Suppose $f : \mathbb{D}\setminus \{0\} \rightarrow \mathbb{C} $ is analytic and $z=0$ is an essential singularity of $f$. Show that the family $\{f_n\} $ defined by
$$ f_n(z) = f \left( \frac{z}{2^n} \right), \quad z \in \mathbb{D}\setminus \{0\} $$
is not normal in $\mathbb{D}\setminus \{0\}$.
My attempt (inspired on this): Assume that $\{f_n\}$ is a normal family in $\mathbb{D}\setminus \{0\}$. So there exists from the sequence $(f_n)_n$ a convergent subsequence $(f_n)_{n_k}$ with limit $\hat{f}$. Because $f$ is analytic, it follows that $\hat{f}$ is analytic on $\mathbb{D}\setminus \{0\}$. We take an annulus $A$ of inner radius $1/3$ and outer radius $1/2$. Then we have that $\hat{f}(A)$ is bounded and because $f_{n_k} \rightarrow f$ as $k \rightarrow \infty$, we have that $f_{n_k} = f(\frac{A}{2^{n_k}})$ is bounded for $k$ large enough. By the Cauchy formula for Laurent series $$ |a_{-j}|  \leq r^j \max_{|z| = r}{|f|}.$$
We choose $r$ as the radius of the circle contained in $A/n_k$ and letting $k\rightarrow \infty$. We then have to show that $|a_{-j}|=0$ for every $j \leq -1$ such that we conclude that $f$ is an analytic function on $\mathbb{D}$ which gives us a contradiction.
But in the last part I got stuck. Because if $k \rightarrow \infty$ then $r \rightarrow 0$ and because $f$ has an essential singularity in $z=0$: $\max_{|z| = r}{|f|}  \rightarrow \infty $. So we have something of the form $0 \cdot \infty$ which is not defined.
Someone who can point out where I went wrong and help me out? Thanks!
 A: You have shown that $\max_{|z| = r_k}{|f|}$ is uniformly bounded for
$$
r_k = \frac{1}{2 \cdot 2^{n_k}}
$$
therefore choosing these radii in
$$
|a_{-j}|  \leq r^j \max_{|z| = r}{|f|}.
$$
implies that $a_{-j} = 0$ for $j \ge 1$.

Alternatively one can argue as follows: $\hat f$ is bounded on the circle $|z| = 1/2$ and $f_{n_k} \to \hat f$  uniformly on that circle. It follows that
$$
 |f_{n_k}(z)| \le M  \text{ for } |z| = \frac 12
$$
with some $M> 0$ and sufficiently large $k$. Translated back to $f$ this means that
$$
 |f(z)| \le M \text{ for } |z| = \frac{1}{2 \cdot 2^{n_k}} \, .
$$
So $|f|$ is bounded by $M$ on those concentric circles. Using the maximum modulus principle it follows that the same estimate holds in the annuli between those circles:
$$
 |f(z)| \le M \text{ for } \frac{1}{2 \cdot 2^{n_{k+1}}} \le |z| \le \frac{1}{2 \cdot 2^{n_k}} \, .
$$
and that implies that $f$ is bounded in a neighborhood of $z=0$.
Using Riemann's theorem it follows that $f$ has a removable singularity at $z=0$.
