Connectivity of Complex Analysis Polynomial Sets | Degree n+1

Mystic998
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Homework Statement



Let p(z) be a polynomial of degree n \geq 1. Show that \left\{z \in \mathbb{C} : \left|p(z)\right| &gt; 1 \right\}[/tex] is connected with connectivity at most n+1.<br /> <br /> <h2>Homework Equations</h2><br /> <br /> A region (connected, open set) considered as a set in the complex plane has finite connectivity n if its complement has n connected components in the extended complex plane.<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> I&#039;m not sure where to start, frankly. Showing the set is connected seems really tricky, though I&#039;m admittedly probably overlooking something really obvious. As for connectivity, I think it has to do with the fact that the complement of the set is \left\{z \in \mathbb{C} : \left|p(z)\right| \leq 1\right\} \cup \left\{\infty\right\} in the extended complex plane. So I think that because the polynomial has at most n roots, any preimage of the first set can have at most n disjoint connected sets mapped to it, then the point at infinity gives you one more connected component. But I&#039;m not sure how to say that rigorously.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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