Does there exist an analytic function that satisfies these properties? Does there exist an analytic function $f:\{z\in\mathbb{C}:0<|z|<1\}\to\mathbb{C}$ such that $\displaystyle\lim_{z\to0}[z^{-3}f^2(z)]=1$?
I'm assuming that there is not such a function, so I've been trying to prove this. Assuming that such a function does exist, there have been a few ideas I've tried. The first was the following:
$\displaystyle\lim_{z\to0}[z^{-3}f^2(z)]=1\implies g:\{z\in\mathbb{C}:0<|z|<1\}\to\mathbb{C}$ by $g(z):=z^{-3}f^2(z)$ has a removable singularity at $z=0$, and so $\displaystyle\lim_{z\to0}z^{-2}f^2(z)=0\implies \displaystyle\lim_{z\to0}z^{-1}f(z)=0$, and then I get stuck.
Another attempt was the following (still assuming, in hopes of contradiction, that such a function $f$ exists):
$\displaystyle\lim_{z\to0}[z^{-3}f^2(z)]=1\implies\lim_{z\to0}[(z^{-1}f^{2/3}(z))^3]=1\implies\lim_{z\to0}[z^{-1}f^{2/3}(z)]=$ some cube root of unity. From this, we can get that $f^{2/3}(z)$ as a removable singularity at $z=0$, and so $\displaystyle\lim_{z\to0}f^{2/3}(z)=0$. Now I know that we define $f^{2/3}(z)=\exp[\frac{2}{3}\log[f(z)]]$ (where $\log$ is a branch of the logarithm), but again I get stuck here. 
I realize that nowhere in either attempt did I use the fact that $f$ is defined on the punctured disk, and maybe using a power series would be helpful? I don't really know. Any help would be appreciated; this is preparation for a qualifying exam and do not feel confident in my analysis skills at all.
 A: If the order of the zero of f at $0$ is two, then you can write a Taylor expansion at zero, $$ f= \sum _{n=0}^{\infty} {a_n z^{n+2}}$$ and factor out $z^2$ so that $$f=z^2(a_0+a_1z+...)$$ and thus $$f^2=z^4(a_0+a_1z+...)^2$$ and then the limit divided by $z^3$ cant be one.
Same goes if the order of zero is higher than two.
A: Hint:
$$\lim_{z\to0}[z^{-3}f^2(z)]=1 \implies f(z)=\frac{1}{z^{3/2}}\frac{1+a_1z+a_2z^2+\cdots}{1+b_1z+b_2z^2+\cdots}$$
A: Assume that such an $f$ exists, and  consider the function
$$g(z):=\left({f(z)\over z}\right)^2\qquad(z\in\dot U)\ ,$$
where $U$ is a sufficiently small disk with center $0\in{\mathbb C}$.
Since $\lim_{z\to0}{g(z)\over z}=1$ it follows that $g$ is in fact analytic in all of $U$, and that there is an analytic function $h:\ U\to{\mathbb C}$ with $$g(z)=z\>h(z)\quad(z\in U),\qquad h(0)=1\ .$$ We may assume that $h(z)\ne0$ in $U$. Since $U$ is simply connected the function $h$ has a well-defined square root $k:\ U\to{\mathbb C}$. We therefore have
$$\left({f(z)\over z}\right)^2=z\bigl(k(z)\bigr)^2\qquad (z\in\dot U)\ ,$$
or
$$z=\left({f(z)\over z\>k(z)}\right)^2=:\bigl(\phi(z)\bigr)^2\qquad (z\in\dot U)\ ,$$
where $\phi$ is analytic in $\dot U$. 
But we all know that there is no analytic square root $\sqrt{z}$ in a full punctured disk around $0$.
