- #1

- 33

- 0

(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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- Thread starter Phillips101
- Start date

- #1

- 33

- 0

(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...)

- #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?

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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- #3

- 33

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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

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

- 33

- 0

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

- #7

- 1

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

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