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Consider the monic polynomial

$$p_{\zeta}(z) = z^n + a_{n-1}(\zeta)z^{n-1} + \dots + a_0(\zeta), $$

where the $a_{i}$'s are continuous functions defined over $\mathbb{C}$. As is well known, the roots of this polynomial depend 'continuously' on the $a_{i}$'s in some sense. However, I was wondering if a stronger statement holds: is it possible to pick continuous functions $f_1, \dots, f_n$, at least locally, such that the roots of $p_{\zeta}(z)$ are exactly $f_1(\zeta), \dots, f_n(\zeta)$, counting multiplicities?

Edit: Note this is easily answered in the affirmative around a $\zeta_0$ such that $p_{\zeta_0}(z)$ has $n$ distinct roots. The real difficulty is in dealing with the case where $p_{\zeta}(z)$ has repeated roots.

user18063
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  • This could be related http://math.stackexchange.com/questions/63196/continuity-of-the-roots-of-a-polynomial-in-terms-of-its-coefficients – PhoemueX Sep 21 '14 at 21:03
  • Isn't this just a combination of the well-known "roots depend continously on the coefficients" and the fact that continuous functions are closed under composition? – Martin Brandenburg Sep 21 '14 at 21:22

1 Answers1

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No, this is not possible, and the example is $z^2-\zeta=0$. Even if you restrict $\zeta$ on the unit circle, there are evidently no continuous functions $z_1(\zeta)$ and $z_2(\zeta)$ defined for all $\zeta$ on the unit circle, and giving the two roots of this equation. Even ONE continuous function $z(\zeta)$ on the unit circle, that satisfies $z^2(\zeta)=\zeta$ evidently does not exist.

Alexandre Eremenko
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