Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $ n \in \mathbb{N} $ and $ a_{0},\ldots,a_{n-1} \in \mathbb{C} $ be constants. By the Fundamental Theorem of Algebra, the polynomial $$ p(z) := z^{n} + \sum_{k=0}^{n-1} a_{k} z^{k} \in \mathbb{C}[z] $$ has $ n $ roots, including multiplicity. If we vary the values of $ a_{0},\ldots,a_{n-1} $, the roots will obviously change, so it seems natural to ask the following question.

Do the $ n $ roots of $ p(z) $ depend on the coefficients in an analytic sort of way? More precisely, can we find holomorphic functions $ r_{1},\ldots,r_{n}: \mathbb{C}^{n} \to \mathbb{C} $ such that $$ z^{n} + \sum_{k=0}^{n-1} a_{k} z^{k} = \prod_{j=1}^{n} [z - {r_{j}}(a_{0},\ldots,a_{n-1})]? $$

The definition of a holomorphic function of several complex variables is given as follows:

Definition Let $ n \in \mathbb{N} $ and $ \Omega \subseteq \mathbb{C}^{n} $ be a domain (i.e., a connected open subset). A function $ f: \Omega \to \mathbb{C} $ is said to be holomorphic if and only if it is holomorphic in the usual sense in each of its $ n $ variables.

The existence of $ r_{1},\ldots,r_{n}: \mathbb{C}^{n} \to \mathbb{C} $ that are continuous seems to be a well-known result (due to Ostrowski, perhaps?), but I am unable to find anything in the literature that is concerned with the holomorphicity of these functions.

Any help would be greatly appreciated. Thank you very much!

share|cite|improve this question
Have you checked $n=2$? ;) – Martin Brandenburg Mar 3 '13 at 3:24
Certainly, if the $a_i$ are such that the roots are all distinct, then the implicit function theorem (and the fact that $P'(z_i)\neq 0$ at any one of the simple roots $z_1,\dots,z_n$) will tell you that the roots move analytically as you perturb the coefficients slightly. This should work more generally for non simple roots if you consider the function that takes a unitary polynomial to its unordered $n$-uple of roots $\in\Bbb C^n/S_n$. – Olivier Bégassat Mar 3 '13 at 3:30
In regards to the continuity of the zeros, in his book Geometry of Polynomials, the earliest citations Marden has for the continuity of roots are H. Weber, Lehrbuch der Algebra, Vol 1, Braunshweig 1895, pp. 132-137 and J. L. Coolidge, The continuity of the roots of an algebraic equation, Ann. of Math. 9 (1908), 116-118. I would not be surprised if it was known before these were published, though. – Antonio Vargas Mar 3 '13 at 3:31
up vote 7 down vote accepted

Such holomorphic functions exist locally as long as you stay away from places where two or more of the roots coincide, but the global situation is more complicated. To get an idea of what happens, consider the case $n=2$, i.e., quadratic equations, and, to make things clearer yet, consider the subspace of $\mathbb C^2$ where $a_1=0$. So you're looking at the equation $z^2+a_0=0$. The solutions, of course, are the two branches of $\sqrt{-a_0}$. They're locally holomorphic away from the origin, but there's a branch point at the origin, and the two solutions (away from the origin) are actually branches of a single, two-valued analytic function.

The general situation is similar, but bigger and in higher dimensions. Away from the locus $D$ where two (or more) of the roots coincide, you have $n$ local holomorphic functions, but these are branches of a single $n$-valued analytic function, with branching along the locus $D$.

share|cite|improve this answer
Professor Blass, do you happen to know of any references that prove the existence of such locally holomorphic functions? Thank you very much for your response. – Haskell Curry Mar 3 '13 at 5:32

To give a little expansion to @Andreas’s answer, let’s examine a little more closely the way the coefficients depend on the roots. Let’s take an $n$-tuple of roots, say $\rho=(\rho_1,\cdots,\rho_n)$ and form the corresponding $n$-tuple whose entries are the coefficients $a=(a_0,\cdots,a_{n-1})$ of the monic polynomial whose roots are the $\rho_i$’s. You have the map $C\colon\rho\mapsto a$, and you can ask what the Jacobian determinant is of this map, call it $J$. Then the fact is that $J^2$ is the discriminant of the polynomial $F(x)=x^n+\sum_0^{n-1}a_{n-i}x^i$, which as you probably know is a polynomial in the $a_i$’s. This fact makes it very clear, via the Inverse Function Theorem, how and when and why the roots depend on the coefficients.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.