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The complex valued function $f(z) = z^n$ has an analytic antiderivative on $\mathbb{C} \setminus \{0 \}$ for every $n$ except for $n=-1$. What is so special about $-1$?

To show why this is such an anomaly, imagine if $z^n$ had an analytic antiderivative on $\mathbb{C} \setminus \{0 \}$ for every $n$ except for $n=3456$. People would demand to know what is so special about $3456$. However, it seems like no one feels the need to explain the anomaly at $n = -1$. What is going on at $-1$?

fdzsfhaS
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  • and why is that? I'm looking for an argument from first principles – fdzsfhaS May 01 '20 at 14:28
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    I think this is fair as a math question (although 'soft'). – Jair Taylor May 01 '20 at 14:28
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    I am not sure what kind of answer you are looking for, but every function meromorphic near the origin has a Laurent series $\sum_{n=-\infty}^\infty a_n z^n$. If you differentiate that, which exponent does not occur in the derivative? – Martin R May 01 '20 at 14:42
  • @DavidWarrenKatz If there were no straight lines in the physical universe every theorem of Eucidean geometry would still be true. The only thing that would change is whether or not Euclidean geometry was a good model of the universe. – John Douma May 01 '20 at 14:45
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    @DavidWarrenKatz a place to start might be that $\frac{1}{z}$ is (up to a constant) the Green's function for $\frac{\partial}{\partial \overline z}$ (cf. https://math.stackexchange.com/questions/833181/greens-function-for-frac-partial-partial-barz) – failedentertainment May 01 '20 at 14:48
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    Wait, are you claiming that $z^{-2}$ has holomorphic antiderivative on an open set containing the origin? – user269711 May 01 '20 at 14:52
  • In complex numbers, the existence of an antiderivative can be characterized by path integrals. Think about taking a path integral around $\frac{1}{z}$. Can that ever be zero? – John Douma May 01 '20 at 15:13
  • I think this question gets to the heart of complex analysis: Cauchy's theorem, Cauchy's integral formula, the residue theorem, etc., and a very intuitive answer would go a long way in making these more understandable. So, please don't close this question, as it is a very valuable one - unless it is a duplicate (which is, sadly, likely.) I'm interested to see the many interpretations of why $z \mapsto z^{-1}$ is so important. I'll try to put an answer to this later, when I have some time. – Jair Taylor May 01 '20 at 15:56
  • @AbhishekShivkumar That is a nice perspective which could be a good answer here with a little elaboration. – Jair Taylor May 01 '20 at 16:03
  • @JairTaylor But, the question doesn't seem well-posed, unless I'm missing something. It's not true for any $n\leq -1$. I think what they mean to say is, "Why does the failure begin at $n=-1$" ($-1$ is not the only integer). – user269711 May 01 '20 at 16:04
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    @user269711 I think the OP meant to (or should have meant to) say that for every integer $n \neq -1$, $z^n$ has an antiderivative $z^{n+1}/(n+1)$ on $\mathbb{C} \backslash {0}$. So $n = -1$ specifically is important, not just any negative integer. – Jair Taylor May 01 '20 at 16:17
  • @JairTaylor But they specifically said an open set containing the origin.. – user269711 May 01 '20 at 16:36
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    @user269711 Sure. My point is that, once you make this correction (removing the origin) the question becomes valid and very interesting. – Jair Taylor May 01 '20 at 16:41
  • @JairTaylor I agree with you on that question being interesting, but I'm saying that it seems like a different question than what they asked. I guess we should wait to see if they update the question to reflect this or not. – user269711 May 01 '20 at 16:43
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    I meant an open punctured neighborhood around $0$. Any easy way to say this? I have changed it to $\mathbb{C} \setminus 0$. – fdzsfhaS May 01 '20 at 17:38
  • Interesting video on the topic: https://www.youtube.com/watch?v=umKYBn2q6e4 – JMP May 02 '20 at 11:09
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    I am looking for an intuitive understanding, not a proof. The linked already answered question is not what I am looking for. Can someone please open this question back up? – fdzsfhaS May 06 '20 at 18:52
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    I disagree that this is a duplicate. This is a soft question, the second one is a hard question. David is not literally asking for a proof that $1/z$ has no antiderivative. –  May 13 '20 at 02:55

3 Answers3

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When you integrate around the circle, the little bits $x^k dx$ rotate through $k+1$ multiples of $2\pi$, so they cancel each other out unless $k=-1$.
We want the integral from $x$ to $y$ along one path to be the same as the integral along another path. Then the integral is just a function of $x$ and $y$. Imagine putting a path from $y$ to $w$ on the end of that, then the integral from $x$ to $w$ equals the integral from $x$ to $y$ plus the integral from $y$ to $w$. So the integral is now a function $F(y)-F(x)$.
If two different paths from $x$ to $y$ have the same integral, then the loop from $x$ to $y$ along the first path, and back along the second path, has to be $F(y)-F(x)+F(x)-F(y)=0$. To repeat, the integral along a closed loop has to be zero.
Suppose a loop can be shrunk to zero, and f(x) is bounded inside the loop. Break up the area into $N^2$ tiny areas of width $O(1/N)$. The integral around the large loop equals the sum of the integrals around all the small areas because the inner paths cancel out. Now we want a condition that makes the integral around a small area to be $O(N^{-3})$. From memory, that condition is the 'Cauchy Riemann' equations, which is the condition the function f is a function of $z$ and not its conjugate $\overline{z}$.
Now the integral is the sum of $N^2$ integrals, each $O(N^{-3})$, so it is $O(1/N)$. Let $N$ be very large, and so the integral is zero. So: The integral around a loop is zero, when $f(z)$ is bounded within the loop, and $f(z)$ does not involve the conjugate.
The only thing left in this case is an integral around zero. This boils down to a single integral along the unit circle. So $z=e^{i\theta}$,$dz=e^{i\theta}id\theta$, $z^kdz = ie^{(k+1)\theta}d\theta$, and the result follows.

Empy2
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What's special about $n=-1$ is that it's the only exponent such that $(az)^n\,d(az)=z^n\,dz$. Here's why that's important:

Keep in mind that if $F(z)$ is analytic in a domain $\Omega$ (such as $\mathbb{C}\setminus\{0\}$), then $F(b)-F(a)=\int_a^bF'(z)\,dz$, where the $\int_a^b$ is understood as a contour integral along any path entirely within $\Omega$ connecting two points $a$ and $b$ in $\Omega$. Now suppose $f(z)=1/z$ had an antiderivative in the domain $\mathbb{C}\setminus\{0\}$, i.e., there were an analytic function $F(z)$ such that $F'(z)=1/z$. Then the function

$$L(z)=F(z)-F(1)=\int_1^z{d\omega\over\omega}$$

is also analytic in the domain $\mathbb{C}\setminus\{0\}$. But we now see that, for any two nonzero complex numbers $a$ and $b$, we have

$$L(ab)=\int_1^{ab}{d\omega\over\omega}=\int_1^a{d\omega\over\omega}+\int_a^{ab}{d\omega\over\omega}=\int_1^a{d\omega\over\omega}+\int_1^b{d(a\omega)\over a\omega}=\int_1^a{d\omega\over\omega}+\int_1^b{d\omega\over\omega}=L(a)+L(b)$$

(i.e., we've just confirmed that the function $L$ behaves like a logarithm). In particular, if $\zeta$ is an $N$th root of unity, we have

$$NL(\zeta)=L(\zeta^N)=L(1)=\int_1^1{d\omega\over\omega}=0$$

Since the totality of all $N$th roots of unity (for all $N\in\mathbb{N}$) is dense on the unit circle, this shows that the analytic function $L(z)$ is identically $0$ on the unit circle, and that implies $L(z)$ is identically $0$ on its entire domain, which implies $L'(z)$ is also identically $0$, which is a contradiction, since $L'(z)=1/z$.

Barry Cipra
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The reason is that the power rule for derivatives won't work: we get division by zero. Of course, as we know from calculus, this is the birth of the natural log. Well, the complex logarithm is analogous.

It is rather amazing that when defining, in the real case, $\ln x=\int_1^x1/t\operatorname dt$, it turns out that we get the inverse of the function $e^x$, where $e=\lim_{x\to\infty}(1+1/x)^x$.

So, in other words, $\ln x=\log_ex$.

Of course, in the complex case we need a branch of $\log$. But branches of $\ln$ are defined in terms of the real $\ln$ via $\ln z=\ln|z|+2\pi k+i \operatorname{arg}z$.

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    what is a cubby corner? Perhaps there is some other formulation of the power rule that does not brake down at -1 – fdzsfhaS May 06 '20 at 18:46