Complex Analysis: brach of the square root

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SUMMARY

The discussion focuses on the existence of a branch of the square root of a quadratic polynomial function f(z) with distinct roots z_1 and z_2 within a domain D. It is established that if a branch z of the square root of f exists in D, then neither z_1 nor z_2 can belong to D. The conversation also raises the question of whether this statement holds true if f has a double root, prompting further exploration into the implications of root multiplicity on branch existence.

PREREQUISITES
  • Understanding of complex analysis concepts, particularly branch cuts.
  • Knowledge of quadratic polynomial functions and their properties.
  • Familiarity with continuity and the definition of branches of complex functions.
  • Basic understanding of root multiplicity in polynomials.
NEXT STEPS
  • Study the concept of branch cuts in complex analysis.
  • Explore the properties of quadratic polynomials and their roots.
  • Learn about the continuity of complex functions and implications for branch existence.
  • Investigate the effects of root multiplicity on the behavior of complex functions.
USEFUL FOR

Students and researchers in mathematics, particularly those studying complex analysis, as well as educators looking for insights into teaching polynomial functions and their properties.

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Homework Statement


Let [itex]f[/itex] be a quadratic polynomial function of [itex]z[/itex] with two different roots [itex]z_1[/itex] and [itex]z_2[/itex]. Given that a branch [itex]z[/itex] of the square root of [itex]f[/itex] exists in a domain [itex]D[/itex], demonstrate that neither [itex]z_1[/itex] nor [itex]z_2[/itex] can belong to [itex]D[/itex]. If [itex]f[/itex] had a double root, would the analogous statement be true?

Homework Equations


We say that a branch [itex]g(z)[/itex] of the [itex]p^{th}[/itex] root of [itex]z[/itex] exists on [itex]D[/itex] if [itex]g(z)[/itex] is continuous and [itex]g(z)^{p}=z[/itex] for every [itex]z \in D[/itex].

The Attempt at a Solution


I am really not sure at all how to begin this proof. I would appreciate a nudge to get started. Thank you.
 
Last edited:
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Use
Code: 
"[itex]1+1=2[/itex]"
instead of
Code: 
"[tex]1+1=2[/tex]"
 
Thanks; I wondered what was up with that...
 

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