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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!

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  • $\begingroup$ 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 $|z|$. $\endgroup$ May 17, 2017 at 17:18

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To compute the integral, you need to compute the residue at infinity. If you increase the radius, since the integrand is holomorphic outside of the unit disk, you don't change the value of the integral.

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 $\int h(z) dz = -i\pi$

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  • $\begingroup$ 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 $(z w)^\alpha \neq z^\alpha w^\alpha$ $\endgroup$
    – Shashi
    May 27, 2017 at 14:52
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    $\begingroup$ 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 $h$ $\endgroup$
    – mercio
    May 27, 2017 at 15:56
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    $\begingroup$ 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. $\endgroup$
    – mercio
    May 27, 2017 at 16:18

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