Intuition for $\int_Cz^ndz$ for $n=-1, n\neq -1$

Question

In the course I'm taking in complex analysis we've been taught that $\int_C\frac{1}{z}dz=2\pi i$ (where $C\subset\mathbb{C}$ is the unit circle), but however, $\int_Cz^ndz=0$ for any integer $n$ different from $-1$. I know that this difference stems from the fact that for $n\neq -1$, $z^n$ has an anti-derivative on $\mathbb{C}-\left\{0\right\}$, but for $\frac{1}{z}$ it doesn't work because $log$ has all this branches business.

However, what I'm looking for is a more "visualized" and intuitive way to "feel" the difference. This relates to having intuitive understanding of the meaning of contour integral, which I unfortunately can't say I have yet. I hope some answers to this particular question will help me develop one.

So, just to clarify, what I'm asking is: Is there a particular feature of the way $1/z$ "looks" (perhaps it's 4-dimentional graph or its action on the complex plane or the Riemann sphere) that gives away the fact that its contour integral is "special"?

I'd like to hear your ways of thinking and imagining regrading these subjects. Thanks!

Answer 1

I don't know if this helps for complex, but for real analysis if you write

$\int_1^x t^{-1+c}dt =\dfrac{x^c-1}{c} $ and let $c \to 0$, you get ln as the limit.

Answer 2

(The following is an explanation in terms of line integrals, as desired.)

When the integer $k$ is ${\ne0}$ the integral ${1\over 2\pi}\int_0^{2\pi}e^{ikt}\>dt$ computes the average complex value of the point $t\mapsto z:=e^{ikt}$ doing $k\ne0$ turns around $S^1$ with constant velocity. This average value is obviously $=0$. If $k=0$ then $e^{ikt}=1$ for all $t$; hence ${1\over 2\pi}\int_0^{2\pi}e^{ikt}\>dt=1$ in this case.

Now the integral $\int_C z^n\>dz$ expands to $$\int_C z^n\>dz=\int_0^{2\pi} e^{i nt}\>i e^{it}\>dt=2\pi i \cdot{1\over 2\pi}\int_0^{2\pi} e^{i (n+1)t}\>dt\ .$$ This is $=0$ if $n+1\ne0$, and is $=2\pi i$ if $n=-1$.

(Note that $z^n=e^{int}$ $(0\leq t\leq 2\pi)$ has average $0$ when $n\ne0$, but the $dz$ adds a complex weight to $z^n$, so that when $n=-1$ we no longer have cancellation.)