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I am trying to understand example VI on the wikipedia page http://en.wikipedia.org/wiki/Methods_of_contour_integration,

but one particular point has mystified me for hours. After it is shown that the line integral representing the left and right circles vanish as the radii get small, there is a skipped step that concludes that

$$\oint_{C}\frac{f(z)}{5-z} dz = (1-i)\int_{0}^{3}\frac{x^\frac{3}{4}(1-x)^{\frac{1}{4}}}{5-x}dx.$$

I understand that this is related to the fact that as the radii approach zero, the horizontal curves approach the branch cut and takes on different values, but I do not understand how to arrive at $1$ and $-i$ respectively.

I have read other examples of this such as the following link:

Contour integration using the residue at infinity

but there is consistently enough detail missing when discussing the branch point to leave me confused at my current level of knowledge in the subject.

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  • $\begingroup$ Are you wondering where the "i" and "" in "$\mathrm{i} r^{3/4}(3-r)^{1/4}$" and "$r^{3/4}(3-r)^{1/4}$" are coming from? ... or how they become "-iI" and "I", respectively? $\endgroup$ Mar 19, 2014 at 6:24
  • $\begingroup$ The former. I don't know where the $i$ and the $1$ come from, I understand why it becomes $-I$ and $I$. $\endgroup$
    – JessicaK
    Mar 19, 2014 at 6:29

1 Answer 1

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For $z^{3/4}$ the chosen branch of the logarithm admits arguments of zero and nearby, so this will contribute: Set $z \rightarrow r \mathrm{e}^{\mathrm{i}\theta}$. Then on $[0,3]$ we find $z^{3/4} \rightarrow r^{3/4}\mathrm{e}^{\mathrm{i}3\theta/4} \xrightarrow[\theta \rightarrow 0]{}r^{3/4} \cdot 1$.

However, this can't work for $(3-z)^{1/4}$ since we only allow arguments $\geq 0$ or $\leq 2\pi$. Notice we have chosen to pay attention to $\arg(3-z)$ so the angles run "backwards" from the normal order, as indicated in the diagram with the "$2\pi$" and "$0$" seemingly in the wrong places around the loop at $z=3$. For the upper branch, $\arg(3-z)$ is near $2\pi$ and we compute $(3-r)^{1/4}\exp(\frac{1}{4}\cdot \mathrm{i} \cdot 2\pi) = (3-r)^{1/4}\cdot \mathrm{i}$. For the lower branch, $\arg(3-z)$ is near $0 = 0\pi$ and we compute $(3-r)^{1/4}\exp(\frac{1}{4}\cdot \mathrm{i} \cdot 0\pi) = (3-r)^{1/4} \cdot 1$. We're literally plugging the $r \mathrm{e}^{\mathrm{i}\theta}$ into the branch choices, moving out the $\log(3-r)$ because $3-r$ is real and seeing what the $\frac{1}{4}$ does to the $\mathrm{i}\theta$ that didn't get moved out.

Not using the shortcut in the first paragraph, we would compute $r^{3/4}\exp(\frac{3}{4} \cdot \mathrm{i} \cdot (+0\pi)) = r^{3/4}\cdot 1$ and $r^{3/4}\exp(\frac{3}{4} \cdot \mathrm{i} \cdot (-0\pi)) = r^{3/4}\cdot 1$ by the same method as in the second paragraph.

Taking the products of the upper part with the upper part and the lower part with the lower part gives the stated results.

I'm pretty sure that the article's "$r^\frac{3}{4}\exp(3\frac{0\pi \mathrm{i}}{4})$" is a typo. Or the "$r^\frac{3}{4}\exp(\frac{0\pi \mathrm{i}}{4})$" is. Either they should both bear the "$\frac{3}{4}$" from the power of $z$ from which they come or neither should. Perhaps the stray "3" should be a minus sign to indicate small negative arguments. I'm not sure. Either way, the result is dominated by the zero in the numerator, so the result is indistinguishable.

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  • $\begingroup$ This explanation motivates the choice of the second branch cut much more now. I totally misinterpreted the image until you explained the angle running backwards. Everything is crystal clear now, thank you. $\endgroup$
    – JessicaK
    Mar 19, 2014 at 7:41
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    $\begingroup$ @JessicaK: Yeah. The "-z" causing the angles to run backwards bit me some years ago. I'll never forget it now... :-) $\endgroup$ Mar 19, 2014 at 8:48
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    $\begingroup$ That used to confuse me as well. But it's not so much that the angle is running backwards as cutting out where $(3-z)$ is real and positive. And it's a little less confusing if you let $ \displaystyle f(z) = \frac{z^{3/4} (z-3)^{1/4}}{5-z} $ and use $0 < \arg(z), \arg(z-3) \le 2 \pi$. This also has the benefit of forcing $f(z)$ to be real valued on the real axis to the right of $z=1$, removing any ambiguities about the residue at infinity. $\endgroup$ Mar 20, 2014 at 15:56

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