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

Phillips101
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...)

Count Iblis
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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Phillips101
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'?

Count Iblis
Yes, it seems that this doesn't work. I'll take a look later at this again.

Count Iblis
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.

Phillips101
Ah! I forgot about that double zero. Very much appreciated

aiuphei
So how can we apply Rouche to this? Which pair of functions do we choose?