given an entire function $f(z)$, find and classify the singular points of the function $f(\frac{1}{z})$.
my take:
$f(z)$ is entire -> $f(z)$ = $\sum_{n=0}^\infty a_n z^n$
plugging in $\frac{1}{z}$ -> $f(\frac{1}{z})$ = $\sum_{n=0}^\infty a_n z^{-n}$
we define $g(z) = f(\frac{1}{z}) = \sum_{n=-\infty}^0 a_{-n}z^n$
$f(z)$ - polynomial $lim_{z\to0} f(1/z) =\infty $ - z=0 is a pole
$f(z)$ is constant $lim_{z\to0} f(1/z) = Const $ - z=0 is a removeable singularity
$f(z) $ is not a polynomial then the laurent expansion is of the form : $f(z)$ = $\sum_{n=0}^\infty a_n z^n$ we get infinite negative exponents when $z= \frac{1}{z}$ and so z=0 is an essential singularity.