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I have the domain $\mathbb{C} \backslash [0,1]$ and want to show that $$\int_\gamma \frac{1}{z(z-1)}dz = 0$$ for all closed curves $\gamma$. I want to accomplish this by explicitly finding an antiderivative.

I did a partial fraction decomposition and got $$f(z) = \frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}$$

Now $$ F(z) =\text{log}(z-1) - \text{log}(z)$$ would be an antiderivative but I don't feel comfortable with the complex logarithm yet. Is my $F(z)$ well-defined on my whole domain? If not, how do I find a well-defined antiderivative?

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  • $\begingroup$ That is at best you can do (but your $F$ is still not defined on the whole domain). You have to use the fact that $\gamma$ does not pass throught $[0,1]$. $\endgroup$ – user99914 May 15 '15 at 7:22
  • $\begingroup$ @KajHansen: Unless I am mistaken, the Cauchy-Goursat theorem applies to simply-connected domains only. Otherwise you could apply it to $1/z$ on $\Bbb C \setminus \{ 0 \}$ . $\endgroup$ – Martin R May 15 '15 at 7:44
  • $\begingroup$ @MartinR, oops. You are correct. $\endgroup$ – Kaj Hansen May 15 '15 at 7:46
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The expression $$ F(z)=\log(z−1)−\log(z) $$ is not well-defined on your domain, because neither $\log(z−1)$ nor $\log(z)$ are holomorphic in $\mathbb{C} \backslash [0,1]$. One can probably solve that by choosing the arguments of the logarithms carefully, but it becomes much easier if the right-hand side is rewritten as $\log \frac{z-1}{z}$.

The Möbius transformation $T(z) = \dfrac{z-1}{z}$ maps $\mathbb{\hat C} \backslash [0,1]$ conformally onto $\mathbb{C} \backslash (-\infty,0]$, so you can define $$ F(z) = \log \frac{z-1}{z} $$ where $$\log w = \log|w| + i \arg w \quad (-\pi < \arg w < \pi) $$ is a holomorphic branch of the logarithm on $\mathbb{C} \backslash (-\infty,0]$.

Then $F$ is an "explicit" antiderivate of $$ \frac{1}{z-1} - \frac{1}{z} $$ This can either be verified directly, or you argue as follows: In any disc $D \subset \mathbb{C} \backslash [0,1]$ $$ F(z) = \log(z-1) - \log z + C $$ for some holomorphic branch of $\log(z-1)$ and $ \log z$ in $D$, and the result follows by differentiation.

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    $\begingroup$ I have tried to answer the question "I want to accomplish this by explicitly finding an antiderivative.". Of course there are other and simpler arguments why the integral is zero. $\endgroup$ – Martin R May 15 '15 at 8:29
  • $\begingroup$ Thanks, I think this answers my question. Just to curious: how would you do it without the antiderivative? Is there another way besides using the residue theorem (because I don't know this yet)? $\endgroup$ – Marc May 17 '15 at 14:50
  • $\begingroup$ @Marc: I don't know which is the best or appropriate way to do it without residues, sorry. You could substitute $z = 1/t$, then you get an integral of a holomorphic function in a simply-connected domain. $\endgroup$ – Martin R May 17 '15 at 15:00
  • $\begingroup$ Can you expand a bit on this? Transforming the function / domain in such a way that we can apply Cauchy's theorem sounds interesting, but I don't understand it from your brief comment. (Maybe make another answer?) $\endgroup$ – Marc May 17 '15 at 18:38
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The solution depends on your closed curve, which poles it does contain in the interior of γ. Since you excluded [0,1] any closed curve will not contain both the poles, or it will contain both and can't have $γ(t)=0$ or $γ(t)=1$. (Curve must have a finite length)

Lets consider:\begin{align}\oint_\gamma f(z)\, dz\end{align} where $f(z)=\frac{1}{z(z-1)}$

Theorem: \begin{align}\oint_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^n \operatorname{I}(\gamma, a_k) \operatorname{Res}( f, a_k )\end{align}

In this specific case $n=2$ and $a_1=0,a_2=1$

Since a close curve will contain both of them, or none of them, then:

\begin{align}\operatorname{I}(\gamma, a_1)=\operatorname{I}(\gamma, a_2)\end{align}

Therefore: \begin{align}\oint_\gamma f(z)\, dz = 2\pi i \operatorname{I}(\gamma, a_1) \sum_{k=1}^2 \operatorname{Res}( f, a_k )\end{align}

Theorem 2: Pole order = 1 $\Rightarrow$ \begin{align}Res(f,z_i) = \lim_{z\to z_i} \frac{z-z_i}{f(z)}\end{align} At $z=0$,\begin{align}Res(f,0) = \lim_{z\to 0} \frac{z}{z(z-1)} = \lim_{z\to 0} \frac{1}{z-1}=-1\end{align} At $z=1$ \begin{align}Res(f,1) = \lim_{z\to 1} \frac{z-1}{z(z-1)}=1 \end{align}

Therefore:

\begin{align}\int_\gamma \frac{1}{z(z-1)}dz = 2\pi i \operatorname{I}(\gamma, a_1) \sum_{k=1}^2 \operatorname{Res}( f, a_k )=2\pi i \operatorname{I}(\gamma, a_1) (1-1)=0\end{align}

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    $\begingroup$ Latex is not difficult to learn actually... You can click "edit" in the OP's question and you'll see a lot of code you need. $\endgroup$ – user99914 May 15 '15 at 8:00
  • $\begingroup$ And that is the reason you down vote an absolutely correct answer? $\endgroup$ – ntarki May 15 '15 at 8:04
  • $\begingroup$ Your answer is correct, but is of very low quality. (Yes I downvoted your answer). Note not only that your picture is a bit hard to see, but also that the information in the picture cannot be searched through the web. $\endgroup$ – user99914 May 15 '15 at 8:09
  • $\begingroup$ Don't you have to take the winding number into account? How would you show that the winding number of $\gamma$ with respect to $z=0$ is the same as the winding number with respect to $z=1$, so that the residues cancel? (Perhaps I am overlooking something simple.) $\endgroup$ – Martin R May 15 '15 at 8:39
  • $\begingroup$ I will fix the answer to account for that and write it in latex. $\endgroup$ – ntarki May 15 '15 at 8:42
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Another possible argument (in response to your above comment): With the substitution $$ z = \frac 1w \, ,\quad dz = -\frac{dw}{w^2} $$ you get $$ \int_\gamma \frac{1}{z(z-1)} \,dz = \int_{\gamma'} \frac{-1}{\frac 1w(\frac 1w-1)} \,\frac{dw}{w^2} = \int_{\gamma'} \frac{1}{w-1} \, dw $$ where $\gamma'$ is a closed curve in $D = \Bbb C \setminus [1, \infty) $. $D$ is simply-connected and $1/(w-1)$ holomorphic in $D$. It follows from Cauchy's integral theorem that the integral is zero.

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  • $\begingroup$ I won't change the accepted answer because the other one fits my question better, but this is beautiful. Thank you very much! $\endgroup$ – Marc May 17 '15 at 22:33
  • $\begingroup$ I tried to work this out in detail and got the problem that there's no $z \in \mathbb{C} \backslash [0,1]$ such that $w=0$ (we'd have to use $z=\infty$). So it seems to me that we can't include $0$ in the domain after substitution. This would imply the loss of the simply-connected property which is necessary for Cauchy's theorem. $\endgroup$ – Marc May 18 '15 at 17:13
  • $\begingroup$ @Marc: I don' think that you actually need the fact that infinity is mapped to zero. You only need that $\gamma'$ is a closed curve in $D = \Bbb C \setminus [1, \infty) $, $D$ is simply-connected and $1/(w-1)$ holomorphic in $D$. Therefore the integral on the rhs is zero. $\endgroup$ – Martin R May 18 '15 at 17:17
  • $\begingroup$ My reasoning is this: if $0$ isn't in $D$, $D$ isn't simply connected. I don't see how $0$ can be in $D$ because no $z \in \mathbb{C}$ satisfies the equation $0 = \frac{1}{z}$. $\endgroup$ – Marc May 18 '15 at 18:55
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    $\begingroup$ @Marc: No. I have defined $D := \Bbb C \setminus [1, \infty)$. And $\gamma'$ happens to be a curve in $D$, that's all. - Or, put in another way: $\int_{\gamma'} \frac{1}{w-1} \, dw = 0 $ for *all* closed curves in $D$, in particular for those which are the image of a $\gamma$ in $\mathbb{C} \backslash [0,1]$ under $z \to 1/z$ . $\endgroup$ – Martin R May 19 '15 at 22:22

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