Given a laurent series, how do we show it doesnt have a pole? Ive been given a Laurent series and told to show that it doesn't have a pole at z = 0, I tried finding the function the Laurent series represents but that is too complex is there any other way to show the series doesn't have a pole at a certain point?
$\sum_{n=0}^{\infty} \frac{z^n}{-2^n} + \sum_{n=0}^{\infty} \frac{n}{z^n}$
 A: In effect, having a pole of order $m$ is the same thing as the Laurent expansion
$$\sum_{n=0}^\infty a_n(z-z_0)^n + \sum_{n=1}^\infty \frac {b_n} {(z-z_0)^{n}}$$
having the nonzero coefficient $b_m$ but every coefficient after $m$ is zero.
In your example, we have a nonzero coefficeint $b_1 = 1$ and nonzero coefficient $b_2= 2$ and so on.  In fact, the coefficients for the terms with negative exponent $(z-z_0)^{-n}$ have infinitely many nonzero coefficients.  (Note: here $z_0 = 0$.)
Therefore the point $z=0$ is not a pole but is in fact an essential singularity.  In fact this is often the definition of an essential singularity.
[Edit: @robjohn is right, I should have paid attention to the domain of convergence.  Since this series does not converge in any neighborhood of $z=0$ then we cannot read off the pole at $z=0$.  Strictly speaking the given series is not defined interior to $|z|<1$ and it makes no sense to talk about singularities or other properties inside this region.  So strictly speaking, this explains why the given series has no pole at the point $z=0$.]
