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?

  • $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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

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}


\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}

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.