Solving Complex Function: Find Singularity of sin(sqrtZ)/Sqrt(Z)

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Homework Help Overview

The discussion revolves around identifying and classifying the singularity of the function sin(sqrtZ)/Sqrt(Z). Participants are exploring the nature of singularities in complex functions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of L'Hôpital's rule and the method of determining the order of the numerator and denominator separately. Questions arise about how to find the order of a function at a point and the implications of that order for classifying the singularity.

Discussion Status

There is an ongoing exploration of the concept of order in relation to singularities. Some participants have provided insights into determining the order of functions, while others express uncertainty about the classification of singularities based on the order found.

Contextual Notes

Participants are navigating the definitions and implications of singularities, particularly in the context of complex analysis, and are working within the constraints of the homework problem without providing direct solutions.

mkbh_10
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Homework Statement



Locate & name the singularity of the function sin(sqrtZ)/Sqrt(Z) ?

Homework Equations





The Attempt at a Solution



At z= 0 i gives 0/0 form so should i apply L hospital's rule & then proceed ?
 
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The is no need to consider the fraction as an entire entity, instead, one can separately calculate the order of the numerator and denominator independently and then combine them to find the order of the quotient.

Hence, start by determining the order of the numerator and denominator separately.
 
determining the order of the numerator and denominator ?
 
One can determine the order of a function, at a point, by finding the order of the derivative which is non-vanishing at that point. For example, the function,

f(x) = x^2

Has order 2 at x=0 since,

f(0)=0 \;\;,\;\;f^\prime(0) = 0 \;\;,\;\;f^{\prime\prime}(0)=2\neq0

Do you follow?
 
Last edited:
The above function has Order =1 at z= 0 , then ?
 
mkbh_10 said:
The above function has Order =1 at z= 0 , then ?
Correct. So, if a function has a singularity of order one what type of singularity is it?
 
i dn't know
 
mkbh_10 said:
i dn't know
A function with a positive order, at a given point, means that the Laurent series of the function at that point has no principle part, which means the singularity is ________.
 

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