How would I count the zeros of zsin(z)-1 in a complex disc?

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Discussion Overview

The discussion revolves around counting the zeros of the function zsin(z) - 1 within a complex disc defined by {z: |z|<(n+1/2)pi}. Participants are exploring the application of Rouche's theorem to establish the number of roots and are particularly interested in confirming that all roots of zsin(z) = 1 are real.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant seeks assistance in applying Rouche's theorem to the function zsin(z) - 1.
  • Another participant proposes a function g(z) = -R^2 sin(pz)/z to compare with f(z) = z sin(z) - 1, questioning if the inequality |f(z) - g(z)| < |f(z)| holds on the boundary of the disc.
  • A different participant points out that the proposed g(z) has an odd number of roots within the disc, while f(z) is believed to have an even number, suggesting that the approach may not be valid.
  • One participant acknowledges the oversight regarding the double zero of g(z) at z = 0, which f(z) lacks.
  • Another participant expresses uncertainty about how to properly apply Rouche's theorem and asks for clarification on the choice of functions.

Areas of Agreement / Disagreement

Participants do not reach consensus on the appropriate functions to use with Rouche's theorem, and there are competing views regarding the validity of the proposed approaches. The discussion remains unresolved.

Contextual Notes

There are limitations regarding the assumptions made about the number of roots for the functions involved, and the selection of parameters in the proposed functions is not fully clarified.

Phillips101
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The disc in question is {z: |z|<(n+1/2)pi}. I can't figure out how to apply Rouche to this. Any help would be appreciated.

(This is in the context of showing all roots of zsin(z)=1 are real. I counted the zeros of zsin(z)-1 on the real axis and got 2n+2, and now I hope to get the same answer via the disc...)
 
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Does this work:

f(z) = z sin(z) - 1

g(z) = -R^2 sin(pz)/z

where R is the radius of the disc and p some carefully chosen number very close to 1 and you try to show that:

|f(z) - g(z)| < |f(z)| on the boundary of the disk?
 
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At first glance, your g(z) has an odd number of roots within the disc, whereas f(z) would have an even number (I'm fairly sure of this, since f(z)+1 is even and the first peak's height is greater than 1 - look at a plot of it on wolframalpha or something if you're not convinced). So, I don't think your function can work.

What was your thinking behind selecting it? What exactly did you mean by 'carefully chosen p'?
 
Yes, it seems that this doesn't work. I'll take a look later at this again.
 
f(z) = z sin(z) - 1

g(z) = z sin(z)

should work if you take into account that g(z) has a double zero at z = 0 that f doesn't have.
 
Ah! I forgot about that double zero. Very much appreciated
 
So how can we apply Rouche to this? Which pair of functions do we choose?
 

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