Integral: $\int_{|z|\mathop=2}\sqrt{z^4-z}\,dz$ - finding Residues / Laurent Series

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)

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}

• 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? May 13 '17 at 2:55
• 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 $-\frac12$. May 13 '17 at 3:10
• Ah yes, I can't believe I didn't spot that! Thanks May 13 '17 at 3:30
• You're welcome! It was my pleasure John. -Mark May 13 '17 at 3:33
• $\texttt{Mathematica}$ yields a 'monster solution'. May 13 '17 at 23:14

$\newcommand{\bbx}{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$ \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,

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

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

• @JohnDoe Thanks. It avoids the paraphernalia of branch cuts. May 14 '17 at 1:53