With regards to utilizing Cauchy's Integral Theorem for integration over closed contours:
https://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem
In particular the result that $\int_\gamma f(z)\,dz = 0$ for closed paths $\gamma$ and holomorphic functions $f(z)$.
It is stated (and shown algebraically) in numerous tutorials and texts that $f(z)=z^{-1}$ violates the conditions for the above result to hold. But in all elaborations given to this statement it has not been clear to me why this is not also true for $z^{-2}$ (or indeed any $z^{n}$ for integers $n$ smaller than -1).
For example, in the Wikipedia article linked, it states:
"The Cauchy integral theorem does not apply here since $f(z)=\frac1 z$ is not defined (and certainly not holomorphic) at $z=0$."
Which I believe is also true for any other $f(z) = \frac 1 {z^n}$.
Alternatively, a set of notes from the Columbia's Complex Analysis course state:
$1/z$ is analytic in the region $\mathbb C − \{ 0 \} = \{z ∈ \mathbb C > : z \neq 0\}$, but this region is not simply connected.
Again believe this also applies to other $f(z) = \frac 1 {z^n}$.
Could someone please clarify what exact condition $\frac 1 z$ violates that $\frac 1 {z^n}$ does not?
P.S. Some background - I am not a mathematician by training but trying to teach myself for applied research, so answers that favor intuition are greatly appreciated!
All the integral powers $z^n$ have an antiderivative $\frac{1}{n+1} z^{n+1}$ on $\mathbb{C}\backslash\{0\}$, except $z^{-1}$. Now, if $F'(z) = f(z) $ then $$\int_{\gamma} f(z) dz = F(\textrm{end point} ) - F(\textrm{initial point} )$$ so in particular, for a closed curve $$\int_{\gamma} f(z) dz = 0$$ if $f$ has an antiderivative on the domain.