How to Find and Classify the Singular Point for f(z)?

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Homework Help Overview

The discussion revolves around finding and classifying the singular points of the function f(z) = 1/(sin z - sin a), where a is an arbitrary real constant. Participants explore the nature of singularities arising from the equation sin z = sin a.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the infinite number of singularities related to the equation sin z = sin a and consider the implications of rewriting the denominator. There is a suggestion to examine the relationship between z and a, with some uncertainty about how to define singularities in this context.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have offered mathematical transformations and relationships, while others express uncertainty about how to proceed with defining singularities. No explicit consensus has been reached.

Contextual Notes

Participants are navigating the complexities of the problem, including the implications of the arbitrary constant a and the resulting singularities. There is a noted lack of clarity regarding the relationship between z and a, which may affect the classification of singular points.

Swati Jain
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Homework Statement


Find and classify the singular point for
f(z) = 1/ ( sin z - sin a)
Where a is an arbitrary real constant.

Homework Equations


f(z) = 1/ ( sin z - sin a)
Where a is an arbitrary real constant.

The Attempt at a Solution


There will be infinite number of singularities of sin z = sin a
Put z' = z-a
Denominator can be written as sin z - sin a = sin ( z'+ a) - sin a
= sin a cos z' + cos a sin z' - sin a = sin a ( cos z' - 1) + cos a sin z'
= sin a ( -1/2! Z'^ 2 + 1/4! Z'^ 4 + ...) + cos a ( z' - z'^ 3/3! +...)
= cos a z' - 1/2! Sin a z'^2 -...
Now how to define a and singularities ??
 
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What if you look at the solution set to the problem:
##\sin a = \sin z##.
 
RUber said:
What if you look at the solution set to the problem:
##\sin a = \sin z##.
I thought in this way too.. But I could not find how to find relation between z and a. It may be z = a + 2n pi, But I am not sure. And further I don"t understand that how to find singularity in this case. Thanks for your response.
 
Maybe you can use the formula for a difference of sines:

##\sin z - \sin a = 2\sin(\frac{z-a} {2})\cos(\frac{z+a} {2})##
 

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