Analyticity of $\dfrac{1}{z}$ vs. $\dfrac{1}{z^2}$ I am learning complex analysis on my own. I am familiar with the theorems, and I am able to compute by hand and get correct results. But there is something that escapes me.
What is the criteria for analyticity?
For example:
derivative if $\dfrac{1}{z}=-\dfrac{1}{z^2}$
derivative of $\dfrac{1}{z^2}=-\dfrac{2}{z^3}$
Both derivatives are undefined at $z=0$. Yet closed curve integral of $1/z !=0$, while $1/z^2$ does.
Cauchy integral theorem states that closed curve integrals of all analytic functions $=0.$
http://mathworld.wolfram.com/ResidueTheorem.html
It states that (integral of) all the terms in a Laurent series besides $a_{-1} =0$ because of the Cauchy integral theorem.
I get the same correct results, computing these by hand. But I would like to understand, why all $1/z^n$ where $n=-1$ closed curve integrals $=0$ due to Cauchy integral theorem. They all have poles at $z=0$. And Cauchy integral theorem is suposed to apply only to curves not containg any poles(in the area they enclose).
I hope I was clear enough in my wording, if not, please say so. 
Post-Acceptance-Edit:
Confusion arose, at their (sites') statement that Cauchys' theorem is the reason for the integral being zero, when that in fact is not true, as confirmed by other users.

If you change variables to $w=\frac{1}{z}$ then you end up with $\int z^{-n}dz$ becoming $-\int w^n \frac{dw}{w^2}=-\int w^{n-2}dw$ and hence, for each $n\geq 2$ this is analytic again.
