I am trying to develop an intuition for the difference between $\oint_C 1/z^2 dz = 0$ and $\oint_C 1/z \,dz = 2\pi i$.

The classical proof of Cauchy's integral theorem seems to be to use Green's theorem along with the Cauchy-Riemann equations. This however, requires a simply connected domain, which $1/z^2$ doesn't have. So if $1/z$ and $1/z^2$ both have a pole at $0$, why do they behave differently when integrating around a closed curve?

I know how to compute the integrals by parameterizing a circle of radius $1$ around $0$, so this is not what I am looking for. Terence Tao has an interesting explanation which I have a hard time wrapping my head around, so perhaps someone could explain starting from here?

Another way to view Cauchy’s theorem is an assertion that every (continuously) differentiable function has an antiderivative. This is true infinitesimally (because $f(z_0)+f'(z_0)(z-z_0)$ has an antiderivative of $f(z_0)(z-z_0) + \frac{1}{2} f'(z_0) (z-z_0)^2)$, and it propagates to be true locally (by summing up and estimating the errors), and then (assuming no topological obstructions, such as poles) it is true globally. In one dimension, the corresponding statement is that every continuous function has an antiderivative (i.e. the fundamental theorem of calculus). In two dimensions, one needs an extra order of control on the function (continuous differentiability rather than just continuity) because one needs one better order of control on the error term to compensate for the extra dimension (dividing a non-infinitesimal two-dimensional region into infinitesimal ones requires many more pieces than for a one-dimensional region, thus allowing many more errors to accumulate.)

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    $\begingroup$ (I'm not sure this answers your question, so I'll leave it as a comment.) It may be helpful to note that the antiderivatives of $1/z,1/z^2$ are $\ln z,-1/z$ respectively. For the latter of those, the antiderivative is a single-valued function of $z$ on $\mathbb{C}\setminus \{0\}$. By contrast, $\ln z$ is a multi-valued function. $\endgroup$ May 2, 2017 at 16:13
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    $\begingroup$ Another observation is that $\frac{1/a}{z-a}-\frac{1/a}{z+a}\to \frac{2}{z^2}$ as $a\to 0$. Hence we can view $\frac{2}{z^2}$ as the coalescence of two poles with equal but opposite residues, so a contour integral which is large enough to enclose both of them must vanish. $\endgroup$ May 2, 2017 at 16:22
  • $\begingroup$ Following your first comment, then I should modify my question a little to: why are we justified in taking the antiderivative for $1/z^2$ but not for $1/z$? I thought it had to do with the domain being simply-connected, but it seems not! $\endgroup$
    – samlaf
    May 3, 2017 at 14:23

2 Answers 2


Let us consider two types of functions: functions with antiderivative and holomorphic functions.

Theorem: if a function $f(z)$ on a domain $D$ has an antiderivative $F(z)$ with the same domain, and if $C$ is a curve in $D$, then $$ \int_C f(z)dz = F(\beta) - F(\alpha), $$ where $\alpha$ is the starting point of $C$ and $\beta$ is the ending point of $C$.

A function with an antiderivative is holomorphic (though we need more than Cauchy's integral theorem to see this). On the contrary, not all holomorphic functions have antiderivative. However:

Theorem: a holomorphic function $f(z)$ has an antiderivative locally. That is, for any point $\alpha$ in the domain $D$ of $f(z)$, there exists a small open disk $B$ with center $\alpha$ such that $f(z)$ has an antiderivative on $B$.

This is just another form of Cauchy's integral theorem. For a holomorphic function $f(z)$ and a path $C$ in the domain $D$ of $f(z)$, we can move path $C$ locally without changing the value $\int_C f(z)dz$ by the first theorem.

The function $1/z$ does not have an antiderivative whose domain is $\mathbb{C}-\{0\}$. To me this is just a fact: $1/z$ is a holomorphic function on $\mathbb{C}-\{0\}$ but not a function with an antiderivative on $\mathbb{C}-\{0\}$.

(Note that $\log z$ is an antiderivative of $1/z$ on $\mathbb{C} - \mathbb{R}_{\leq 0}$, or any domain which does not contain a loop around $0$.)


Let $n$ be a integer and $n \neq 1$. It is well-known that (if $C$ is a closed curve containing the origin in its interior) that $\int_{C}\frac{1}{z^n} \, dz$ is zero. For our proof, the idea will be $C$ can be deformed to a circle $C'$ containing the origin in it's interior. We can (in fact) let $C'$ be the circle $|z|=1$. Now, for any point on the unit circle, $z=e^{i \theta}$ where $\theta$ is real. And $dz=iz d\theta$. So $\int_{C'}\frac{1}{z^n} \, dz$ is $\int_{0}^{2\pi}i\frac{1}{e^{in\theta}}e^{i\theta} \, d\theta$. This is $i\int_{0}^{2\pi}e^{-(n-1)i\theta} \, d\theta$. As $n \neq 1$, we can express $i\int_{0}^{2\pi}e^{-(n-1)i\theta} \, d\theta$ as $i(\int_{0}^{2\pi}\cos(n-1)\theta \, d\theta-i\int_{0}^{2\pi}\sin(n-1)\theta)\, d\theta$. It should be easy to show $\int_{0}^{2\pi}\cos(n-1)\theta \, d\theta$ and $\int_{0}^{2\pi}\sin(n-1)\theta \, d\theta$ are each zero. For example, if $n=2$, then $\int_{0}^{2\pi}\cos \theta \, d\theta$ is zero follows from a sketch of the graph of cosine.


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