- #1

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(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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In summary, the conversation discusses the use of Rouche's Theorem to show that all roots of zsin(z)=1 are real. The suggested approach involves choosing carefully selected functions f(z) and g(z) and showing that |f(z) - g(z)| < |f(z)| on the boundary of a chosen disc. However, upon further examination, it is determined that this approach will not work due to the double zero in g(z). The conversation ends with the question of how to apply Rouche's Theorem in this case.

- #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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- #2

- 1,863

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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?

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

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What was your thinking behind selecting it? What exactly did you mean by 'carefully chosen p'?

- #4

- 1,863

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

- #5

- 1,863

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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

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

- #7

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

In order to determine the number of zeros of zsin(z)-1 in a complex disc, you can use the argument principle or Rouché's theorem. These methods involve analyzing the number of roots of the function within a given region based on the number of times the argument changes.

No, the quadratic formula is only applicable for finding the roots of a quadratic equation with real coefficients. Since zsin(z)-1 is a complex function, other methods must be used to find its zeros.

No, according to the fundamental theorem of algebra, a polynomial function can only have a finite number of roots in the complex plane. Therefore, zsin(z)-1 can only have a finite number of zeros in a given complex disc.

Yes, the zeros of zsin(z)-1 can be located at non-real complex numbers. This is because the sine function can take on complex values, and when multiplied by z, the function can have zeros at non-real complex numbers.

The location of the zeros of zsin(z)-1 can change as the value of z varies in the complex disc. This is because the function is highly sensitive to small changes in z and the zeros can move around in the complex plane as z varies. However, the total number of zeros will remain the same within the given disc.

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