How do i get sin(z)/2i = 1/4(e - 1/e) ? Resides at Poles question?

  • Thread starter Thread starter laura_a
  • Start date Start date
  • Tags Tags
    Poles
Click For Summary
SUMMARY

The discussion focuses on evaluating the residue of the function f(z) = sin(z)/(z^2+1) at its poles, specifically at z = i. The residue is calculated using the formula Res[f(z), z_0] = lim(z→z_0) (z-z_0) f(z), leading to Res[f(z), i] = 1/(2i). The key transformation involves using the identity sin(i) = (e^(-1) - e)/(2i), which simplifies to 1/4(e - 1/e). This transformation is crucial for understanding the relationship between the sine function and exponential functions in complex analysis.

PREREQUISITES
  • Complex analysis, specifically residue theory
  • Understanding of poles and residues in complex functions
  • Familiarity with the sine function's exponential representation
  • Basic calculus for evaluating limits
NEXT STEPS
  • Study the properties of residues in complex analysis
  • Learn about the exponential form of trigonometric functions
  • Explore the application of the residue theorem in contour integration
  • Practice evaluating residues for various complex functions
USEFUL FOR

Students of complex analysis, mathematicians focusing on residue theory, and anyone looking to deepen their understanding of the relationship between trigonometric and exponential functions in the context of complex variables.

laura_a
Messages
64
Reaction score
0

Homework Statement



Find Poles & eval resides at those points for

f(z) = sin(z)/(z^2+1)



Homework Equations



I'm using Res f(z) = [g^m-1 (z_0)] / (m-1)!

The Attempt at a Solution



So I worked out the f(z) = 1/(z-i) so g(z) = sin(z) /(z+i) for my first one did the sub for z=i and got sin(i)/2i

But the working out (i'm working on past papers so I have the worked solutions, just not very detailed) The next line shows;

= 1/4(e-(1/e))

What one Earth don't I know about sin that gets to this point??
 
Last edited:
Physics news on Phys.org
For simple pole at [tex]z_0[/tex], the residue of f(z) is given by
[tex]\text{Res}\left[f(z), z_0\right]= \lim_{z\rightarrow z_0} (z-z_0) f(z)[/tex]

eg.
[tex]\text{Res}\left[\frac{1}{(z^2+1)}, i \right]= \lim_{z\rightarrow i} \frac{z-i}{z^2+1} =\lim_{z\rightarrow i} \frac{1}{z+i}=\frac{1}{2i}[/tex]

now, looks like you have got it right so far with sin(i)/2i, all it is need to get that next line is just by using
[tex]\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}[/tex]
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K