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The integral expression for the complex logarithm is defined by: $$\int_{\gamma} \frac{1}{z}\,dz$$ where $\gamma$ represents a rectifiable path in $\mathbb{C}\setminus\{0\}$. The above integral defines $\text{log}(z)$, which has a branch cut emanating from $0$. And this is the source of my confusion when it comes to the calculation of the residue at $0$. I mean, are we supposed to insert a branch cut at zero and then consider a circular path around the branch ?

Edit

I will rephrase the above question. The logarithm is defined as: $$\text{log}(z)=\text{log}|z| + i(\theta + 2\pi k)$$ Here, the integer $k$ defines the Riemann sheet, in other words how many times you transverse the branch cut emanating from $0$ in the positive sense. In order to cross the branch cut we must however trace around the zero. The question is then: does the factor $2\pi i k$ in fact represent the residue contribution ?

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  • $\begingroup$ It is not clear what exactly you mean with your opening sentence "the integral expression for the complex logarithm is defined by: $\int_{\gamma} \frac{1}{z}\,dz$". It is incorrect to state that this integral gives a complete definition of the logarithm, if that is what you mean. $\endgroup$ – Ben Grossmann Jul 17 at 9:47
  • $\begingroup$ @BenGrossmann Hello Ben. Thank you for the answer and comments. Could you tell me what is missing from the definition ? Is there chat room that we can discuss ? $\endgroup$ – user91411 Jul 17 at 9:49
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    $\begingroup$ I don't think there's any problem with discussing the question here in the comments. The problem with your definition is that the result of $\int_\gamma \frac 1z$ depends on which path from $1$ to $w \in \Bbb C$ is taken, so you have not defined a unique value for $\log(w)$. If you want to think of $\log$ as a multivalued function, then presumably the log should be defined to be the set of all possible results. The multivalued function does not itself have a branch cut, but the associated single-valued functions do. $\endgroup$ – Ben Grossmann Jul 17 at 9:57
  • $\begingroup$ @BenGrossmann I have edited the question. $\endgroup$ – user91411 Jul 17 at 10:21
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The reason that it is interesting to calculate the residue of $\frac 1z$ at $0$ (or equivalently, the counterclockwise integral $\oint_{|z| = 1} \frac 1z dz$) is that it cannot be (directly) calculated using an antiderivative.

If $\gamma$ is a curve in $\Bbb C$ that starts at $a \in \Bbb C$ and ends at $b \in \Bbb C$ and if there exists a function $F(z)$ such that $F$ is differentiable at all points in $\gamma$ with $f(z) = \frac d{dz}F(z)$, then we have $$ \int_\gamma f(z)\,dz = F(b) - F(a). $$ It follows that if $\gamma$ is a closed contour (so that $a = b$), then we have $\int_\gamma f(z)\,dz = F(a) - F(a) = 0$. In other words: if we know that there is an anti-derivative of $f$ that is globally defined along the contour $|z| = 1$, then it necessarily follows that its reside at $0$ is $0$.


That said, we can use the antiderivative to compute the residue if we split the desired integral into parts. Let $\gamma_1$ denote the path along $|z| = 1$ from $1$ to $-1$, and let $\gamma_2$ denote the path along $|z| = 1$ from $-1$ to $1$, both taken counterclockwise. We have $$ \oint_{|z| = 1}\frac 1{z}\,dz = \int_{\gamma_1} \frac 1zdz + \int_{\gamma_2} \frac 1z dz. $$ We now consider two different antiderivatives for $\frac 1z$ corresponding to different branch cuts. Define $\log^1,\log^2$ such that $$ \log^1(e^{i \theta}) = i\theta, \quad \theta \in [-\pi/2,3 \pi/2);\\ \log^2(e^{i \theta}) = i\theta, \quad \theta \in [\pi/2, 5\pi/2). $$ We then have $$ \int_{\gamma_1} \frac 1zdz = \log^1(-1) - \log^1(0) = \pi i - 0 = \pi i,\\ \int_{\gamma_2} \frac 1z dz = \log^2(0) - \log^2(-1) = 2 \pi - \pi i = \pi i. $$


Regarding your edited question: your definition $$ \log(re^{i\theta}) = \{\log r + i(\theta + 2 \pi k) : k \in \Bbb Z\} $$ is consistent with the definition $$ \log(z) = \left\{\int_{\gamma}\frac 1z \,dz : \gamma \text{ is a contour from } 1\ \text{to}\ z\right\}. $$ It is indeed the case that the multiple of $k$ corresponds to the contribution of the residue at $z = 0$.

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  • $\begingroup$ Does that mean the notion of branch cut when calculating the residue does not apply ? $\endgroup$ – user91411 Jul 17 at 9:35
  • $\begingroup$ @user91411 I'm not sure what you mean by "doesn't apply" here. I've added something, maybe that clarifies the situation. $\endgroup$ – Ben Grossmann Jul 17 at 9:40
  • $\begingroup$ @user91411 More typically, this integral is calculated by parameterizing the contour with $\gamma:[0,1] \to \Bbb C$, so that $$ \int_\gamma \frac 1z \,dz = \int_0^1 \frac{1}{\gamma(t)} \frac{d \gamma}{dt}\,dt. $$ $\endgroup$ – Ben Grossmann Jul 17 at 9:41
  • $\begingroup$ Looking at the second part of your answer it looks as if we need to define two branches and two paths that don't cross the cuts to make a full circle. I was wondering whether this can be done using a single cut around which we use a single path crossing the cut. For instance, for the second branch in the answer if we consider the path to be full circle with initial and final points on the opposite sides of the branch, we again obtain $2\pi i$. $\endgroup$ – user91411 Jul 19 at 19:28
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    $\begingroup$ @user91411 As I say in my edited answer, the answer to that is yes. $\endgroup$ – Ben Grossmann Jul 19 at 21:35

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