Connectivity of Complex Analysis Polynomial Sets | Degree n+1

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SUMMARY

The discussion centers on the connectivity of the set defined by the polynomial p(z) of degree n ≥ 1, specifically the set {z ∈ ℂ : |p(z)| > 1}. It is established that this set is connected with a connectivity of at most n+1. The reasoning involves analyzing the complement set {z ∈ ℂ : |p(z)| ≤ 1} and its relationship to the roots of the polynomial, which contribute to the number of connected components in the extended complex plane.

PREREQUISITES
  • Understanding of complex analysis concepts, particularly polynomial functions.
  • Familiarity with the topology of the complex plane and connected sets.
  • Knowledge of the concept of connectivity in topological spaces.
  • Experience with the extended complex plane and its components.
NEXT STEPS
  • Study the properties of polynomial functions in complex analysis.
  • Learn about the topology of the complex plane, focusing on connectedness and open sets.
  • Explore the concept of connectivity in topological spaces, including definitions and examples.
  • Investigate the implications of the extended complex plane and its role in complex analysis.
USEFUL FOR

Students and researchers in complex analysis, mathematicians focusing on polynomial behavior, and anyone interested in the topological properties of sets in the complex plane.

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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