# When is the square root of a polynomial holomorphic? And how do you take the contour integral?

I am asked to show that if $p(z)$ is a polynomial such that all the roots satisfy $|z| < R$, then for $|z| > R$ there is an analytic/holomorphic function $h(z)$ such that $(h(z))^2 = p(z)$ when $p(z)$ is of even degree.

I was tempted to think that this was trivially true, but I suppose it depends on the exponential definition of powers, specifically that $$z^a = e^{\frac{1}{2}log(z)}$$ where are appropriate branch of log is chosen. The evenness of $p$ then comes in handy since it ensures that if we cross over the branch cut, the total jump in the argument of the exponential is $2ni\pi$ for some integer $n$, thus ensuring that the square root of $p$ is well defined for $|z| > R$ where crossing over the branch cut ensures that every linear factor of $p$ also crosses it. This would explain why the conditions of $|z| > R$ and evenness were given.

Is this the right way to think about it?

In the next part of the question I am asked to compute $$\int_C \sqrt{z^4 -z} \; dz$$ where $C = \{z \in \mathbb C \;|\; |z| = 2\}$

It suggests that I work out the Laurent Series in order to do this question. I am not sure how to work out the Laurent expansion here, nor am I sure how else to do it.

I have thought that perhaps since the contour contains all the roots of $p(z) = z^4 - z$ and thus $h(z) = \sqrt{p(z)}$, and since all the roots are branch points, i.e. points that cannot have a defined argument, then I can apply the residue theorem and get:

$$\int _ C \sqrt{z^4 - z} \;dz = \sum^{4}_{i=1} I(C,a_i)Res(h,a_i)$$ where $(a_1,a_2,a_3,a_4) = (0,1,w,w^2)$ and $w$ is a strictly complex cube root of unity.

But even if this were true, I don't know how to calculate the residues.

I feel like I'm missing something really simple. Any help would be appreciated, thank you!

• Hint: $\sqrt{z^4-z} = z^2\sqrt{1-\frac{1}{z^3}}$ for $|z| > 1$ and $\approx z^2 ( 1 - \frac12 z^{-3} + O(z^{-6}))$ for large $|z|$. May 17, 2017 at 17:18

If you pick the branch $h(z)$ whose value at $z=2$ is $h(2) = +\sqrt{14}$, then $h$ has an asymptotic development at infinity :
$h(z) = z^2 -\frac 12 z^{-1} + O(z^{-4})$
Now, as the radius increases, the integral of the $O(z^{-4})$ term converges to $0$, and since it's independant of the radius, it is $0$.
Moreover, $\int z^2 dz = 0$, and $\int z^{-1} dz = 2i\pi$, and so $\int h(z) dz = -i\pi$
• I think that you have used: $\sqrt[]{z^4-z} = \sqrt[]{z^4}\sqrt[]{1-\frac{1}{z^3}}$ to get the Laurent series for $\vert z\vert >1$. But I don't understand why it is allowed to say that: $\sqrt[]{z^4-z} = \sqrt[]{z^4}\sqrt[]{1-\frac{1}{z^3}}$ in this case. Could you explain that, please? Because I know in general that $(z w)^\alpha \neq z^\alpha w^\alpha$ May 27, 2017 at 14:52
• If you have a holomorphic function $g(z)$ on the open unit disk such that $g(z)^2 = 1+z$, then picking $h(z) = z^2 g(-z^{-3})$ defines a holomorphic function on the exterior of the unit disk satisfying $h(z)^2 = z^4 - z$. And if you know the Taylor development for such a $g$ it gives you a Laurent series for the corresponding $h$ May 27, 2017 at 15:56
• alternatively you can simply say $(h(z)-z^2+\frac 12 z^{-1})(h(z)+z^2-\frac 12 z^{-1}) = \ldots = - \frac 14 z^{-2} = O(z^{-2})$. Since the two factors differ by a $\theta(z^2)$, if one factor is "small" (a $o(z^2))$ then the other is "large" (a $\theta(z^2)$), and so that small factor is actually a $O(z^{-4})$, so very small. May 27, 2017 at 16:18