Are the following theorems' converses also true? Suppose $f(z) → \infty$ as $z → z_0$, so there's an isolated singularity at $z = z_0$.  Ultimately I intend to show that there is a pole at $z_0$?  I realize this has been asked before, but my question is different because I am asking specifically about the approach outlined below and whether or not the converse of the following two thereoms are true.
I know that I can do this by showing that $z=z_0$ is neither a removable singularity nor an essential singularity.  However, I'm not positive why it's not either of those.
I know that if $f$ is bounded in a deleted neighborhood of an isolated singularity, then the singularity is removable.  But I don't know if the converse is true. Is it?  Also I know that if $f$ has an isolated singularity at $z_0$ and if lim$_{z→z_0}(z-z_0)f(z) = 0$, then the singularity is removable.  But again I don't know if the converse is true. Is it?
To show it's not an essential singularity, I can just use the Casorati-Weierstrass Theorem and show that if $D$ is a deleted neighborhood of $z_0$, then the range $R = {{f(z):z\in D}}$ is not dense in the complex plane.  But how would I show that this is true ?
 A: If $f$ has a removable singularity at $z_0$, then we can extend $f$ to $z_0$ to take a value $f(z_0)=a$ so that it is an analytic function at $z_0$.  In particular, it is continuous at $z_0$, so by definition of continuity, for any $\epsilon>0$ there is a neighborhood of $z_0$ on which $|f(z)-a|<\epsilon$.  In particular, $|f(z)|< |a|+\epsilon$ is bounded on such a neighborhood.  Similarly, by continuity $\lim_{z\to z_0}(z-z_0)f(z)=(z_0-z_0)a=0$.
Also, if $f(z)\to\infty$ as $z\to z_0$, then for any $N$, there is a deleted neighborhood of $z_0$ on which $|f(z)|>N$, and then the image of that neighborhood cannot be dense as long as $N>0$ (because the image is contained in the closed set $\{w:|w|\geq N\}$).
But a simpler approach to this problem (which does not require the full classification of isolated singularities) is just to consider the function $g(z)=1/f(z)$.  Since $f(z)\to \infty$ as $z\to z_0$, $g(z)\to 0$ as $z\to z_0$, so $g$ has a removable singularity at $z_0$ with value $0$.  Filling in that singularity, we can write $f(z)=1/g(z)$, where $1$ and $g(z)$ are both analytic functions at $z_0$, with $g(z_0)=0$ and $1(z_0)\neq 0$.  Thus $f$ has a pole at $z_0$.
A: What if
$f(z)
=\exp(\frac1{|z-z_0|})
$?
There weren't any
restrictions on $f$
that I could see.
