Intuitive reason why contour integral around origin of $z^n$ is 0 unless n = -1

Question

The contour integral of an analytic function around a point is determined entirely by the $\frac{1}{z}$ term in a function's Laurent series. In particular,

$\int_{\gamma} z^n dz = \begin{cases} 2 \pi i & n = -1 \\ 0 & n \ne -1 \end{cases} $

Where $\gamma$ is any contour wrapping once around the origin.

I understand how to do the calculation and derive this result, but as this result is important to complex analysis I want to understand it better. Is there an intuitive reason why this is true?

For example, $z^{-3}$ and $z^{-1}$ qualitatively look similar, so why do the integrals turn out qualitatively differently?

Or to ask differently, why does $z^n$ not have an antiderivative when $n = -1$

Answer 1

In very simple terms: when $z$ describes a curve around the origin, $z^ndz$ also describes a curve around the origin, which it surrounds $n+1$ times. But for $n=-1$ it surrounds the origin zero times, because it is constant !

Below the curves for the image of a shifted circle containing the origin. The black dot corresponds to $n=-1$.

Due to the distribution along the curve, the integrals turn out to vanish.

Answer 2

Deform the contour to the unit circle (which by holomorphicity of $z^m$ on the set $\mathbb{C}/0$ is as good as any other) and set $e^{ix}=z$. $$ I=\int_{\gamma}dz z^m=i\int_0^{2 \pi}e^{i x (m+1)}dx $$ Only for $m=-1$ the resulting integrand won't be a periodic function, whose integral vanishs over a whole period.

As another hint why $m=-1$ might be special, consider $\int dx x^m$ on the real line. Even here things go wild for this value

Answer 3

As you know, $$(z^d)'=dz^{d-1}$$ works for any $d$ but isn't revertible for $d=0$. This must be the reason for the "singularity".

Answer 4

"For example, $z^{-3}$ and $z^{-1}$ qualitatively look similar, so why do the integrals turn out qualitatively differently?"

Answer: $z^{-3}$ has the antiderivative $-z^{-2}/2$ in the domain $\mathbb C\setminus \{0\},$ while $z^{-1}$ has no antiderivative there.

Answer 5

Consider $$\oint_{|z|=r}z^adz$$ for real $a$.

By parametrization, $$=\int^{2\pi}_0 r^ae^{ia\theta}ire^{i\theta}d\theta=r^{a+1}\frac{e^{2\pi ia}-1}{a+1}$$

There is one and the only singularity(removable) at $a=-1$, making $\frac1z$ special.

Imaginary part of $\frac{e^{2\pi ia}-1}{a+1}$ as a function of $a$ is plotted. Maxima $2\pi$ is attained at $a=-1$.