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I wish to calculate the integral $$\int_{|z|\mathop=2}\sqrt{z^4-z}\,dz$$


$$z^4-z=z(z-1)(z-e^{2i\pi/3})(z-e^{-2i\pi/3})$$ We must make branch cuts which go through the branch points. The branch points of $\sqrt{z^4-z}$ are $z=0,1,e^{2i\pi/3},e^{-2i\pi/3}$. All of these lie inside this contour, so I first thought that I must work out the residue at each of these points, and then use the residue theorem. However there is no nice way that I could think of to calculate these residues.

Then it occurred to me that I am not even sure if they exist - these are not poles, they are branch points with branch cuts through them. I then saw that the question has a hint saying to use Laurent Series. Most of the Laurent series I have ever seen have been functions with (sort of) Taylor series, but with some change of variables, or multiplied by some factor of $1/z^n$. The point is - I don't know any (efficient) way of finding the Laurent Series for this function.


My questions: How can I

  1. Define these branch cuts,

  2. Find the different Laurent Series, and find the value of the coefficient of $1/z$, thus finding value of the integral?

Also:

  1. I thought Laurent Series existed within an annulus of the point about which they exist - how is this possible for a function with branch cuts? Is there something I am confusing here? These two ideas I had seem to contradict.

  2. If the Laurent Series do exist, then is it possible to calculate the residues directly? (Also assuming these exist)

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2 Answers 2

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We can cut the plane such that the branch cuts join pairs of the branch points. For example, the cuts are from $0$ to $1$ and from $e^{i2\pi/3}$ and $e^{-i2\pi/3}$.

Note that with this choice of branch cuts, $f(z)=\sqrt{z^4-z}$ is analytic in any annulus $1<|z|<R$. Therefore, we can deform the contour $|z|=2$ to $|z|=R\to \infty$.

The integral of interest becomes

$$\begin{align} \oint_{|z|=2}f(z)\,dz&=\oint_{|z|=R}f(z)\,dz\\\\ &=\oint_{|z|=R} z^2\sqrt{1-\frac{1}{z^3}}\,dz\\\\ &=\oint_{|z|=R} z^2\left(1-\frac{1}{2z^3}+O\left(\frac{1}{z^6}\right)\right)\,dz\\\\ &=\int_0^{2\pi}R^2e^{i2\phi}\left(1-\frac{1}{2R^3e^{i3\phi}}+O\left(\frac{1}{R^6}\right)\right)\,iRe^{i\phi}\,d\phi\\\\ &=-i\pi+O\left(\frac1{R^3}\right)\\\\ &\to -i\pi \,\,\text{as}\,\,R\to \infty \end{align}$$

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  • $\begingroup$ Oh ok I see. So when the hint said to use Laurent Series, it may not have necessarily meant to find Laurent series of $\sqrt{z^4-z}$, and instead it could have meant to do so for something entirely different such as $\sqrt{1-\frac1{z^3}}$? Also, do you know how to answer questions 3 and 4 which I mentioned too? $\endgroup$
    – John Doe
    Commented May 13, 2017 at 2:55
  • $\begingroup$ Yes, you have it. Now, for the third question, since the branch cuts we chose are inside the unit circle, $\sqrt{z^4-1}$ is analytic in any annulus $1<|z|<R$. And for the fourth question, we can see from the expansion $\sqrt{z^4-1}=z^2-\frac1{2z}+O\left(\frac{1}{z^4}\right)$, we see directly that the residue is $-\frac12$. $\endgroup$
    – Mark Viola
    Commented May 13, 2017 at 3:10
  • $\begingroup$ Ah yes, I can't believe I didn't spot that! Thanks $\endgroup$
    – John Doe
    Commented May 13, 2017 at 3:30
  • $\begingroup$ You're welcome! It was my pleasure John. -Mark $\endgroup$
    – Mark Viola
    Commented May 13, 2017 at 3:33
  • $\begingroup$ $\texttt{Mathematica}$ yields a 'monster solution'. $\endgroup$ Commented May 13, 2017 at 23:14
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \oint_{\verts{z} = 2}\root{z^{4} - z}\,\dd z & = \,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\, \oint_{\verts{z} = 1/2}{\root{1 - z^{3}} \over z^{4}}\,\dd z \end{align} There is a pole of order four at $\ds{z = 0}$. The residue at such pole is equal to $\ds{\color{#f00}{-\,{1 \over 2}}}$ which is straightforward evaluated by expanding the square root in powers of $\ds{z}$. Namely,

$\ds{\root{1 - z^{3}} = 1 + \pars{\color{#f00}{-\,{1 \over 2}}}z^{3} + \,\mrm{O}\pars{z^{6}}}$


$$ \oint_{\verts{z} = 2}\root{z^{4} - z}\,\dd z = 2\pi\ic\pars{-\,{1 \over 2}} = \bbx{-\pi\ic} $$

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  • $\begingroup$ @JohnDoe Thanks. It avoids the paraphernalia of branch cuts. $\endgroup$ Commented May 14, 2017 at 1:53

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