Analyticity of $\dfrac{1}{z}$ vs. $\dfrac{1}{z^2}$

Question

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.

Answer 1

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 $n=1$.

Answer 2

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 $n=1$.

Answer 3

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.