# Contour Integration of Square Root with Laurent Series

Recently I've been working on branch cuts of square root functions, and come to problems like this:

Find a single-valued analytic branch $f$ of $\sqrt{z^2+z}$ on the set $\{ z \in \mathbb{C}: |z| >1 \}$ such that $f(2) = -\sqrt{6}$. Evaluate the integral of $f$ around the contour $|z| =2$ (positively oriented) using the Laurent series of the square root.

I can find an $f$: the function $-\sqrt{z}\sqrt{z+1}$ where the square roots are the principal branch of the square roots function.

However, the only example of using Laurent series I've seen (in these notes) for contour integrals like this works by factoring a $z$ out of a square root; i.e. writing $$(z^2+z)^{1/2} = z(1+1/z)^{1/2}$$ and then using the Laurent expansion for the (principal branch) square root $\sqrt{z}$ about $z=1$ to expand $(1+1/z)^{1/2}$.

My question is about the signs here. The approach above assumes that $\sqrt{z^2} = z$, but it seems that in my branch of the square root, I have $\sqrt{z^2} = -z$.

And then the square root I'm left with in the factorization is still defined based on the branch of $\sqrt{z^2+z} \;$ I'm using, so would I use the Laurent expansion for $- \sqrt{z}$ instead?

I understand that if both signs need to be negative, there's no net effect on the result; I'm just trying to understand the details of the computation. Any insight would be appreciated.

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Concretely expand $\sqrt{z^2+z}=-z\sqrt{1+1/z}\$ that verifies $f(2)=-\sqrt{6}$ (the $+$ possibility doesn't apply).

The cut should be chosen between the points $-1$ and $0$ as in your linked file because $f(2)=-\sqrt{6}$ is indicated without additional specification (compare to 'the top side of the cut' in the pdf) so that the function must be smooth at point $z=2$. This point would indeed be on the cut had we chosen the cut outside of $(-1,0)$.

Here is a picture of the argument of your function as considered (the minus adds $\pi$ to the argument) :

note that for $z$ real with $z<-1$ the arguments become equal to 0 $\pmod {2\pi}$

The same kind of problem was handled in this other thread with the cut shown outside the points or inside. Note that there is a link to the handling of the general case at the end (just set a=−1 and b=0 there).

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Alright, thanks. I see how the factorization works here. I still don't see why it works, though. If I'm just given that $f$ is an branch of $\sqrt{z^2+z}$ with $f(2) = -\sqrt{6}$, why does it factor to $-z\sqrt{1+1/z^2}$ (principal branch; $|z| >1$)? I know that this factorization is correct up to sign, and I can just check a value to verify the right sign; but that doesn't tell me why it's correct. –  ec92 Jul 14 '12 at 19:08
@ec92: because there are not many Laurent-series around $0$ corresponding : one with the $+$ sign the other with the $-$ sign. With $\sqrt{x}\sqrt{1+x}$ for example the powers of $x$ will be increasing (and 'half-powered') while the square root $z^2(1+1/z)$ will give us the decreasing integer powers that we want. Other 'adequate' formulations should give the same Laurent expansion at the end! –  Raymond Manzoni Jul 14 '12 at 19:26
@ec92: "Because the Laurent expansion of a function is unique whenever it exists, any expression of this form that actually equals the given function $f(z)$ in some annulus must actually be the Laurent expansion of $f(z)$" Wiki Laurent series. –  Raymond Manzoni Jul 14 '12 at 19:32

A function has a branch point $a$, if the function changes its value when the argument increases by $2 n \pi$, $n$ is an integer. That means, if you replace $z$ by $r e^{i w}$ in your function (change to polar coordinates), then replace $w$ by $w + 2 n \pi$, the function changes its value. If you do this you will see the minus sign ( or the other value of the function, since the function is discontinuous at the branch cut).

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