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.
 A: It is true that 
$$\int_{|z|=1} \frac{1}{z^n} \, dz = 0, \:\: n=2,3,4\dots$$
But the reason is NOT Cauchy's Theorem.  For, as you said, these functions are not analytic in the unit disk.  One reason for the above is that $z^{-n}$ has an analytic primitive (antiderivative) along the unit circle, unless $n=1$.
A: The Laurent series of $\frac1{z^n}$ is $$ 0+\frac 0z+\frac 0{z^2}+\ldots +\frac 0{z^{n-1}}+\frac1{z^n}+\frac 0{z^{n+1}}+\ldots$$
so the only nonzero coefficient is $a_{-n}$. Hence the coefficient $a_{-1}$ that the Cuachy integral measeres is nonzero iff $n=1$.
A: The problem you're having here with your curve integral is that Cauchy Integral theorem requires the function to be analytic on region inside the curve. But these functions are not analytic at origin.
Note also that the converse is not true, just because the integral vanishes doesn't mean that the function is analytic in the region inside as for example the example $f(z) = z^{-2}$
The definition for analyticity is in general that the function is possible to express as a power series (it doesn't have to be a complex function of a complex variable). At the level of basic complex analysis it's sometimes defined as being differentiable, but I'd recommend against it as a definition (as it only works out in that context). 
A: 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.
