Could someone provide an intuitive explanation for contour Integrals? So my core understanding of what a contour integral is may be wrong so the question I'm attempting to ask is what is the complex output value of a contour integral 'defining'.
I understand well enough that for closed paths not containing singularizes, the contour integral will equal zero and that if it does contain singularities the it may produce a complex valued output given in the form of $\displaystyle 2\pi i\ f( a)$ for cases that can be evaluated using the Cauchy integral formula.
I have no trouble evaluating them mathematically but I have zero idea about why it's even useful to compute a contour integral. If not defining an 'area', what is it actually telling us? How do we use that information? Is their a good visual explanation for why a contour integral is non zero for contours around a singularity?
 A: Contour integrals are not areas. If $\gamma:[0,1]\to\mathbb C$ is piecewise differentiable, then:
$$\int_{\gamma} f(z)\,dz =\int_0^1 f(\gamma(t))\gamma’(t)\,dt$$
Note that $\gamma’(t)$ is a complex number, so the multiplication inside the integral is hiding some complexity.
In particular, it’s not simply described as the areas of the real and imaginary parts of $f(z).$ If $f(z)=1$ For all $z,$ you get a simple result of $\gamma(1)-\gamma(0),$ no matter what path you take.
The fundamental reason the integrals around loops are zero when there are no poles is that we are dealing with a very small set of functions, the analytic functions. These are representable (locally) as power series, and we can (locally) find anti-derivatives for them from the power series.
Now, if $F’(z)=f(z)$ for all $z$ in a neighborhood of your curve, $\gamma:[0,1]\to\mathbb C$, the fundamental theorem of calculus can be shown to give:
$$\int_{\gamma} f(z)\,dz =F(\gamma(1))-F(\gamma(0))$$
When $\gamma$ is a loop, then, $\gamma(1)=\gamma(0),$ so this integral is zero.
Now, any analytic series can be written locally as a power series, and thus, locally, has an anti-derivative.
What happens when $\gamma$ goes around $0$ and you integrate $z^n,$ for $n$ an integer? You get an easy anti-derivative for all $n$ other than $n=-1.$ There is no complex anti-derivative for $z^{-1}$ in a neighborhood of a loop about zero.
It takes a lot more rigor, of course, to get the details right, but the underlying question is “why aren’t these loop integrals always zero?” The answer is that sometimes we can’t find an anti-derivative.
In particular, if $f$ has an anti-derivative, it is analytic. But also, it can’t have any nasty simple poles.

Of course, the loop integral can be zero if you loop once and then loop back in the opposite direction. Then $f$ might not have an anti-derivative near all of $\gamma,$ but you can break the integral up locally into multiple loops where it does have an anti-derivative, and the additional “parts” cancel out. That’s why the integral around zero of $\frac{1}{z}$ only depends on the number of times $\gamma$ goes around $0.$
