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

  • #1

Main Question or Discussion Point

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...)
 

Answers and Replies

  • #2
1,838
7
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?
 
Last edited:
  • #3
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'?
 
  • #4
1,838
7
Yes, it seems that this doesn't work. I'll take a look later at this again.
 
  • #5
1,838
7
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.
 
  • #6
Ah! I forgot about that double zero. Very much appreciated
 
  • #7
1
0
So how can we apply Rouche to this? Which pair of functions do we choose?
 

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

  • Last Post
Replies
1
Views
2K
Replies
10
Views
4K
Replies
2
Views
2K
Replies
7
Views
8K
Replies
22
Views
8K
  • Last Post
Replies
13
Views
3K
Replies
6
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
3
Views
3K
Top