My book claims that $1/(1-z)$ has an essential singularity at $z=1$ by writing out the Laurent series and showing that there are infinitly many terms. Why isn't this just a pole of order 1? We can write this as $-1/(z-1)$, then the numerator is a nonvanishing analytic function, and the denominator is of the form $(z-1)$, so it seems like by definition this should be a pole, and not an essential singularity. What am I missing here?
If anyone is curious: the book is the Princeton Review for the GRE Math Subject Test, 4th edition, p. 312-313
Their argument is this. Consider the region $1 < |z| < \infty$ and write
$f(z) = \frac{1}{1-z} = \frac{1}{z(\frac{1}{z}-1)} = -\frac{1}{z} \cdot \frac{1}{z-\frac{1}{z}}$.
Then if $|z| > 1$, then $|1/z| < 1$, so we can expand
$\frac{1}{1-z} = -\frac{1}{z} \cdot \frac{1}{1-\frac{1}{z}}= -\frac{1}{z} (1 + (\frac{1}{z} + \frac{1}{z}^2 + \cdots) = \frac{1}{z} - (\frac{1}{z})^2 - (\frac{1}{z})^3 - \cdots$
for $|z| > 1$. Since this has infinitely many terms in the Laurent series, $z=1$ is an essential singularity.