0

https://en.wikipedia.org/wiki/Rouché%27s_theorem

I got a question regarding the proof of Rouches theorem. The proof our prof gave us tells us that the contour $\partial K$ needs to be partwise smooth, and I can't understand why it would need to be that.

Different proofs seems to have different setups for the countour. If anyone know this proof well, can you explain why the contour needs to be partwise smooth?

uoiu
  • 593
  • 1
    The proof uses the argument principle which requires being able to integrate along the contour $\gamma$. Piecewise smooth is a lazy way to make sure you can do this. Maybe rectifiable is enough, or maybe you need absolute continuity. A specialist in geometric analysis could quickly do better than what I am suggesting. – Charlie Frohman Jan 02 '23 at 13:42
  • Thanks you so much, this answer was exactly what I needed! – uoiu Jan 02 '23 at 13:49

1 Answers1

-1

When you integrate along a curve in C first the ∂ should not contain roots of the function. A more important fact is the homotopy class of that curve(Jordan curve theorem allows you to make those considerations https://en.wikipedia.org/wiki/Jordan_curve_theorem . Usually when you integrate in C you do not require more restrictive conditions you usually find in R. Those considerations lead to the more general hypothesis that in the Rouches Theorem you only need a simple (no self intersections) closed curve. As your link suggest A stronger version of Rouché's theorem was published by Theodor Estermann in 1962.[1] It states: let K ⊂ G K\subset G be a bounded region with continuous boundary ∂ K . So in this case in even more general. P.S You can easily extend the Rouches theorem for meromorphic functions. I hope I provided you more materials to clarify the situation. Have a good day

giorgiokyn23
  • 630