Is this the answer for what is the residue of f(z)=e^z/[(z-1)^2(z+2)]

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

The function f(z) = e^z/[(z-1)^2(z+2)] has poles at z = 1 (order 2) and z = -2 (order 1). The residue at z = 1 is calculated as Res(f,1) = -e/9, while the residue at z = -2 is Res(f,-2) = 1/(9e^2). These results are derived using the residue formula Res(f(z)) = g(p-1)(a) / (p-1)!, where g(z) is appropriately defined for each pole.

PREREQUISITES
  • Complex analysis, specifically residue theory
  • Understanding of poles and their orders
  • Familiarity with the exponential function e^z
  • Knowledge of differentiation and the application of the quotient rule
NEXT STEPS
  • Study the calculation of residues for higher-order poles
  • Learn about Laurent series and their applications in complex analysis
  • Explore the properties of meromorphic functions
  • Review examples of residue calculations in complex integration
USEFUL FOR

This discussion is beneficial for students of complex analysis, particularly those preparing for exams or seeking to deepen their understanding of residue calculations and pole behavior in complex functions.

laura_a
Messages
64
Reaction score
0

Homework Statement


This was an exam question I had today:

Give the location and order of each pole of
f(z)=e^z/[(z-1)^2(z+2)]
and evaluate the reside at these points


Homework Equations



Res(f(z)) = g(p-1)(a) / (p-1)!



The Attempt at a Solution



g(z) = e^z/(z+2)
g'(z) = -e^z/(z+2)^2
let a=1
g'(1) = -e^1/9

So res(f,1) = -e/9

Then I did
g(z) = e^z/(z-1)^2
(this is the part I think I might have stuffed up, because I did use the formula above but I had m=1 for f(z) which was 1/(z+2) so I thought it was right? anyways I got

g(-2) = e^(-2) / (-3)^2 = res(f,-2)=1/(9e^2)

It would mean a lot to me if someone could let me know because I am trying to work out if I should be depressed because I failed or if I can be happy because I passed :)
 
Physics news on Phys.org
At first glance I agree with your first answer for res(f,-2) but not for res(f,1).

<br /> g\left( z \right) = \frac{{e^z }}{{\left( {z + 2} \right)}} \Rightarrow g&#039;\left( z \right) = \frac{{e^z }}{{\left( {z + 2} \right)}} - \frac{{e^z }}{{\left( {z + 2} \right)^2 }}<br />
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
2
Views
2K
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K