maximal injective neighborhoods centered at the zero of a polynomial

173 Views Asked by At

I was working on a particular problem involving the injectivity of a certain polynomial, $p(z) = z^5 + z -1$, $z \in \mathbb{C}$, in which I needed to find a neighborhood around it's real root so that $p(z)$ was one to one in, say $D(\alpha,r)$ where $\alpha$ is the real root of the polynomial, and $r > 0$.

The natural question to me was: How big can we make $r$ so that $p(z)$ is injective in this disk? I figured that there must be a maximal value, but was not sure how to find (other than numerically), how large r would be.

Another question might be to ask what are the largest sets surrounding $p(z)$'s zeros that guarantee injectivity in those regions? It is worthy to note that in this case, all of the zeros of $p(z)$ are of multiplicity $1$, I would assume that it would be impossible to find such a set if there were even multiplicities.

1

There are 1 best solutions below

1
On

For the first part of you question, let us consider the set $C=\{z\in\mathbb{C}\ :\ p'(z)=0\}$. We have $p'(z)=5z^4+1$, hence $$C=\{z\in\mathbb{C}\ :\ z^4=-1/5\}=\{5^{-1/4}\exp(i(1+2k)\pi/4)\ :\ k=0,1,2,3\}$$ Now, set $D=p( C)$ and $D'=p^{-1}(D)$.

It is a standard result that $$p\vert_{\mathbb{C}\setminus D'}:\mathbb{C}\setminus D'\to\mathbb{C}\setminus D$$ is a (holomorphic) covering with (in this case) $5$ preimages for each point.

As a disc is simply connected, for every disc $\Delta\subset\mathbb{C}\setminus D'$ the image $p(\Delta)=\Omega$ is simply connected and so has $5$ disjoint preimages, all biholomorphic, hence all biholomorphic to $\Delta$. When $p$ is restricted to each of those preimages (so also to $\Delta$) it has to be a $1-\textrm{to}-1$ holomorphic covering, i.e. an injective holomorphic function.

On the other hand, as soon as you include in a domain a point where the derivative vanishes, the function has no hopes to be injective on that domain.

So, summing up, you need to find the biggest disc you can fit into $\mathbb{C}\setminus D'$.

In the case at hand, I don't see immediately how to compute $D'$ without some lengthy computation, so I am going to let you do that!

In the general case, you are searching for "the biggest" (if it exists) simply connected, connected open set, containing the root you are interested in, inside $\mathbb{C}\setminus D'$, and yes, if the root is of multiplicity higher than $1$ you have no hopes: locally around it $p$ will look like $t\mapsto t^m$ ($m$ is the multiplicity of the root, t is a local coordinate of the form $t=(z-\alpha)h(z)$ with $h$ non-vanishing around $\alpha$), so it will always be $m-\textrm{to}-1$ in a punctured neighbourhood of $\alpha$.